id
int64
0
14.1k
problem_id
int64
1
1.31k
problem_title
stringclasses
441 values
difficulty
stringclasses
3 values
c_source
stringclasses
441 values
architecture
stringclasses
4 values
optimization
stringclasses
4 values
compiler
stringclasses
8 values
assembly
stringlengths
31
174k
14,100
1,312
Minimum Insertion Steps to Make a String Palindrome
Hard
#include <stdio.h> #include <stdlib.h> #include <string.h> static inline int min(int a, int b) { return a < b ? a : b; } int minInsertions(char * s){ int i, j, len = strlen(s); int *dp = malloc(len * sizeof(int)); memset(dp, 0, len * sizeof(int)); for (i = len - 2; i >= 0; i--) { int left_down = 0; for (j = i + 1; j < len; j++) { int down = dp[j]; if (s[i] == s[j]) { dp[j] = left_down; } else { dp[j] = min(down, dp[j - 1]) + 1; } left_down = down; } } return dp[len - 1]; } int main(int argc, char **argv) { if (argc != 2) { fprintf(stderr, "Usage: ./test s\n"); exit(-1); } printf("%d\n", minInsertions(argv[1])); return 0; }
riscv64
-O0
RISC-V 64 gcc 15.2.0
min: addi sp,sp,-32 sd ra,24(sp) sd s0,16(sp) addi s0,sp,32 mv a5,a0 mv a4,a1 sw a5,-20(s0) mv a5,a4 sw a5,-24(s0) lw a5,-20(s0) mv a2,a5 lw a5,-24(s0) mv a3,a5 sext.w a4,a3 sext.w a5,a2 ble a4,a5,.L2 mv a3,a2 .L2: sext.w a5,a3 mv a0,a5 ld ra,24(sp) ld s0,16(sp) addi sp,sp,32 jr ra minInsertions: addi sp,sp,-64 sd ra,56(sp) sd s0,48(sp) addi s0,sp,64 sd a0,-56(s0) ld a0,-56(s0) call strlen mv a5,a0 sw a5,-32(s0) lw a5,-32(s0) slli a5,a5,2 mv a0,a5 call malloc mv a5,a0 sd a5,-40(s0) lw a5,-32(s0) slli a5,a5,2 mv a2,a5 li a1,0 ld a0,-40(s0) call memset lw a5,-32(s0) addiw a5,a5,-2 sw a5,-20(s0) j .L5 .L10: sw zero,-28(s0) lw a5,-20(s0) addiw a5,a5,1 sw a5,-24(s0) j .L6 .L9: lw a5,-24(s0) slli a5,a5,2 ld a4,-40(s0) add a5,a4,a5 lw a5,0(a5) sw a5,-44(s0) lw a5,-20(s0) ld a4,-56(s0) add a5,a4,a5 lbu a3,0(a5) lw a5,-24(s0) ld a4,-56(s0) add a5,a4,a5 lbu a5,0(a5) mv a4,a3 bne a4,a5,.L7 lw a5,-24(s0) slli a5,a5,2 ld a4,-40(s0) add a5,a4,a5 lw a4,-28(s0) sw a4,0(a5) j .L8 .L7: lw a5,-24(s0) slli a5,a5,2 addi a5,a5,-4 ld a4,-40(s0) add a5,a4,a5 lw a4,0(a5) lw a5,-44(s0) mv a1,a4 mv a0,a5 call min mv a5,a0 mv a3,a5 lw a5,-24(s0) slli a5,a5,2 ld a4,-40(s0) add a5,a4,a5 addiw a4,a3,1 sext.w a4,a4 sw a4,0(a5) .L8: lw a5,-44(s0) sw a5,-28(s0) lw a5,-24(s0) addiw a5,a5,1 sw a5,-24(s0) .L6: lw a5,-24(s0) mv a4,a5 lw a5,-32(s0) sext.w a4,a4 sext.w a5,a5 blt a4,a5,.L9 lw a5,-20(s0) addiw a5,a5,-1 sw a5,-20(s0) .L5: lw a5,-20(s0) sext.w a5,a5 bge a5,zero,.L10 lw a5,-32(s0) slli a5,a5,2 addi a5,a5,-4 ld a4,-40(s0) add a5,a4,a5 lw a5,0(a5) mv a0,a5 ld ra,56(sp) ld s0,48(sp) addi sp,sp,64 jr ra .LC0: .string "Usage: ./test s\n" .LC1: .string "%d\n" main: addi sp,sp,-32 sd ra,24(sp) sd s0,16(sp) addi s0,sp,32 mv a5,a0 sd a1,-32(s0) sw a5,-20(s0) lw a5,-20(s0) sext.w a4,a5 li a5,2 beq a4,a5,.L13 lui a5,%hi(stderr) ld a5,%lo(stderr)(a5) mv a3,a5 li a2,16 li a1,1 lui a5,%hi(.LC0) addi a0,a5,%lo(.LC0) call fwrite li a0,-1 call exit .L13: ld a5,-32(s0) addi a5,a5,8 ld a5,0(a5) mv a0,a5 call minInsertions mv a5,a0 mv a1,a5 lui a5,%hi(.LC1) addi a0,a5,%lo(.LC1) call printf li a5,0 mv a0,a5 ld ra,24(sp) ld s0,16(sp) addi sp,sp,32 jr ra
14,101
1,312
Minimum Insertion Steps to Make a String Palindrome
Hard
#include <stdio.h> #include <stdlib.h> #include <string.h> static inline int min(int a, int b) { return a < b ? a : b; } int minInsertions(char * s){ int i, j, len = strlen(s); int *dp = malloc(len * sizeof(int)); memset(dp, 0, len * sizeof(int)); for (i = len - 2; i >= 0; i--) { int left_down = 0; for (j = i + 1; j < len; j++) { int down = dp[j]; if (s[i] == s[j]) { dp[j] = left_down; } else { dp[j] = min(down, dp[j - 1]) + 1; } left_down = down; } } return dp[len - 1]; } int main(int argc, char **argv) { if (argc != 2) { fprintf(stderr, "Usage: ./test s\n"); exit(-1); } printf("%d\n", minInsertions(argv[1])); return 0; }
riscv64
-O1
RISC-V 64 gcc 15.2.0
minInsertions: addi sp,sp,-48 sd ra,40(sp) sd s0,32(sp) sd s1,24(sp) sd s2,16(sp) sd s3,8(sp) sd s4,0(sp) mv s4,a0 call strlen mv s3,a0 sext.w s0,a0 slli s2,s0,2 mv a0,s2 call malloc mv s1,a0 mv a2,s2 li a1,0 call memset li a5,1 ble s0,a5,.L2 addiw t5,s3,-1 addi t4,s1,-8 add t4,t4,s2 addi a0,s4,-2 add t1,a0,s0 mv t3,t5 j .L7 .L5: addiw a4,a4,1 .L4: sw a4,4(a6) addi a5,a5,4 addi a3,a3,1 beq a3,a7,.L12 .L6: mv a6,a5 mv a4,a1 lw a1,4(a5) lbu a2,0(a3) beq a2,a0,.L4 lw a2,0(a5) mv a4,a2 sext.w a2,a2 ble a2,a1,.L5 mv a4,a1 j .L5 .L12: addiw t3,t3,-1 beq t3,zero,.L2 addi t4,t4,-4 addi t1,t1,-1 .L7: ble s0,t3,.L3 lbu a0,0(t1) addi a3,t1,1 addi a7,t1,2 subw a5,t5,t3 slli a5,a5,32 srli a5,a5,32 add a7,a7,a5 mv a5,t4 li a1,0 j .L6 .L3: addiw t3,t3,-1 addi t4,t4,-4 addi t1,t1,-1 j .L7 .L2: slli s0,s0,2 addi s0,s0,-4 add s1,s1,s0 lw a0,0(s1) ld ra,40(sp) ld s0,32(sp) ld s1,24(sp) ld s2,16(sp) ld s3,8(sp) ld s4,0(sp) addi sp,sp,48 jr ra .LC0: .string "Usage: ./test s\n" .LC1: .string "%d\n" main: addi sp,sp,-16 sd ra,8(sp) li a5,2 bne a0,a5,.L16 ld a0,8(a1) call minInsertions mv a1,a0 lui a0,%hi(.LC1) addi a0,a0,%lo(.LC1) call printf li a0,0 ld ra,8(sp) addi sp,sp,16 jr ra .L16: lui a5,%hi(stderr) ld a3,%lo(stderr)(a5) li a2,16 li a1,1 lui a0,%hi(.LC0) addi a0,a0,%lo(.LC0) call fwrite li a0,-1 call exit
14,102
1,312
Minimum Insertion Steps to Make a String Palindrome
Hard
#include <stdio.h> #include <stdlib.h> #include <string.h> static inline int min(int a, int b) { return a < b ? a : b; } int minInsertions(char * s){ int i, j, len = strlen(s); int *dp = malloc(len * sizeof(int)); memset(dp, 0, len * sizeof(int)); for (i = len - 2; i >= 0; i--) { int left_down = 0; for (j = i + 1; j < len; j++) { int down = dp[j]; if (s[i] == s[j]) { dp[j] = left_down; } else { dp[j] = min(down, dp[j - 1]) + 1; } left_down = down; } } return dp[len - 1]; } int main(int argc, char **argv) { if (argc != 2) { fprintf(stderr, "Usage: ./test s\n"); exit(-1); } printf("%d\n", minInsertions(argv[1])); return 0; }
riscv64
-O2
RISC-V 64 gcc 15.2.0
minInsertions: addi sp,sp,-48 sd ra,40(sp) sd s0,32(sp) sd s1,24(sp) sd s2,16(sp) sd s3,8(sp) mv s3,a0 call strlen sext.w s0,a0 slli s2,s0,2 mv s1,a0 li a1,1 mv a0,s2 call calloc li a5,1 ble s0,a5,.L2 addiw t5,s1,-1 addi t1,s3,-2 addi t4,a0,-8 mv t3,t5 add t1,t1,s0 add t4,t4,s2 ble s0,t3,.L3 .L17: subw a5,t5,t3 slli a5,a5,32 lbu a7,0(t1) srli a5,a5,32 addi a6,t1,2 add a6,a6,a5 addi a3,t1,1 mv a5,t4 li a2,0 .L6: lbu a1,0(a3) mv a4,a2 addi a3,a3,1 lw a2,4(a5) beq a1,a7,.L4 lw a1,0(a5) mv a4,a1 ble a1,a2,.L5 mv a4,a2 .L5: addiw a4,a4,1 .L4: sw a4,4(a5) addi a5,a5,4 bne a3,a6,.L6 addiw t3,t3,-1 beq t3,zero,.L2 .L15: addi t4,t4,-4 addi t1,t1,-1 bgt s0,t3,.L17 .L3: addiw t3,t3,-1 j .L15 .L2: slli s0,s0,2 ld ra,40(sp) add a0,a0,s0 ld s0,32(sp) lw a0,-4(a0) ld s1,24(sp) ld s2,16(sp) ld s3,8(sp) addi sp,sp,48 jr ra .LC0: .string "Usage: ./test s\n" .LC1: .string "%d\n" main: addi sp,sp,-16 sd ra,8(sp) li a5,2 bne a0,a5,.L21 ld a0,8(a1) call minInsertions mv a1,a0 lui a0,%hi(.LC1) addi a0,a0,%lo(.LC1) call printf ld ra,8(sp) li a0,0 addi sp,sp,16 jr ra .L21: lui a5,%hi(stderr) ld a3,%lo(stderr)(a5) lui a0,%hi(.LC0) addi a0,a0,%lo(.LC0) li a2,16 li a1,1 call fwrite li a0,-1 call exit
14,103
1,312
Minimum Insertion Steps to Make a String Palindrome
Hard
#include <stdio.h> #include <stdlib.h> #include <string.h> static inline int min(int a, int b) { return a < b ? a : b; } int minInsertions(char * s){ int i, j, len = strlen(s); int *dp = malloc(len * sizeof(int)); memset(dp, 0, len * sizeof(int)); for (i = len - 2; i >= 0; i--) { int left_down = 0; for (j = i + 1; j < len; j++) { int down = dp[j]; if (s[i] == s[j]) { dp[j] = left_down; } else { dp[j] = min(down, dp[j - 1]) + 1; } left_down = down; } } return dp[len - 1]; } int main(int argc, char **argv) { if (argc != 2) { fprintf(stderr, "Usage: ./test s\n"); exit(-1); } printf("%d\n", minInsertions(argv[1])); return 0; }
riscv64
-O3
RISC-V 64 gcc 15.2.0
minInsertions: addi sp,sp,-48 sd ra,40(sp) sd s0,32(sp) sd s1,24(sp) sd s2,16(sp) sd s3,8(sp) mv s3,a0 call strlen sext.w s0,a0 slli s2,s0,2 mv s1,a0 li a1,1 mv a0,s2 call calloc li a5,1 ble s0,a5,.L2 addiw t5,s1,-1 addi t1,s3,-2 addi t4,a0,-8 mv t3,t5 add t1,t1,s0 add t4,t4,s2 ble s0,t3,.L3 .L20: subw a6,t5,t3 slli a6,a6,32 addi a5,t1,2 srli a6,a6,32 add a6,a6,a5 lbu a7,0(t1) addi a4,t1,1 mv a5,t4 li a1,0 j .L8 .L19: lw a1,0(a5) addi a4,a4,1 mv a3,a1 ble a1,a2,.L5 mv a3,a2 .L5: addiw a3,a3,1 sw a3,4(a5) beq a4,a6,.L6 .L17: addi a5,a5,4 mv a1,a2 .L8: lbu a3,0(a4) lw a2,4(a5) bne a3,a7,.L19 sw a1,4(a5) addi a4,a4,1 bne a6,a4,.L17 .L6: addiw t3,t3,-1 beq t3,zero,.L2 .L16: addi t4,t4,-4 addi t1,t1,-1 bgt s0,t3,.L20 .L3: addiw t3,t3,-1 j .L16 .L2: slli s0,s0,2 ld ra,40(sp) add a0,a0,s0 ld s0,32(sp) lw a0,-4(a0) ld s1,24(sp) ld s2,16(sp) ld s3,8(sp) addi sp,sp,48 jr ra .LC0: .string "Usage: ./test s\n" .LC1: .string "%d\n" main: addi sp,sp,-16 sd ra,8(sp) li a5,2 bne a0,a5,.L24 ld a0,8(a1) call minInsertions mv a1,a0 lui a0,%hi(.LC1) addi a0,a0,%lo(.LC1) call printf ld ra,8(sp) li a0,0 addi sp,sp,16 jr ra .L24: lui a5,%hi(stderr) ld a3,%lo(stderr)(a5) lui a0,%hi(.LC0) addi a0,a0,%lo(.LC0) li a2,16 li a1,1 call fwrite li a0,-1 call exit
14,104
1,312
Minimum Insertion Steps to Make a String Palindrome
Hard
#include <stdio.h> #include <stdlib.h> #include <string.h> static inline int min(int a, int b) { return a < b ? a : b; } int minInsertions(char * s){ int i, j, len = strlen(s); int *dp = malloc(len * sizeof(int)); memset(dp, 0, len * sizeof(int)); for (i = len - 2; i >= 0; i--) { int left_down = 0; for (j = i + 1; j < len; j++) { int down = dp[j]; if (s[i] == s[j]) { dp[j] = left_down; } else { dp[j] = min(down, dp[j - 1]) + 1; } left_down = down; } } return dp[len - 1]; } int main(int argc, char **argv) { if (argc != 2) { fprintf(stderr, "Usage: ./test s\n"); exit(-1); } printf("%d\n", minInsertions(argv[1])); return 0; }
x86-64
-O0
x86-64 clang 21.1.0
minInsertions: push rbp mov rbp, rsp sub rsp, 48 mov qword ptr [rbp - 8], rdi mov rdi, qword ptr [rbp - 8] call strlen@PLT mov dword ptr [rbp - 20], eax movsxd rdi, dword ptr [rbp - 20] shl rdi, 2 call malloc@PLT mov qword ptr [rbp - 32], rax mov rdi, qword ptr [rbp - 32] movsxd rdx, dword ptr [rbp - 20] shl rdx, 2 xor esi, esi call memset@PLT mov eax, dword ptr [rbp - 20] sub eax, 2 mov dword ptr [rbp - 12], eax .LBB0_1: cmp dword ptr [rbp - 12], 0 jl .LBB0_11 mov dword ptr [rbp - 36], 0 mov eax, dword ptr [rbp - 12] add eax, 1 mov dword ptr [rbp - 16], eax .LBB0_3: mov eax, dword ptr [rbp - 16] cmp eax, dword ptr [rbp - 20] jge .LBB0_9 mov rax, qword ptr [rbp - 32] movsxd rcx, dword ptr [rbp - 16] mov eax, dword ptr [rax + 4*rcx] mov dword ptr [rbp - 40], eax mov rax, qword ptr [rbp - 8] movsxd rcx, dword ptr [rbp - 12] movsx eax, byte ptr [rax + rcx] mov rcx, qword ptr [rbp - 8] movsxd rdx, dword ptr [rbp - 16] movsx ecx, byte ptr [rcx + rdx] cmp eax, ecx jne .LBB0_6 mov edx, dword ptr [rbp - 36] mov rax, qword ptr [rbp - 32] movsxd rcx, dword ptr [rbp - 16] mov dword ptr [rax + 4*rcx], edx jmp .LBB0_7 .LBB0_6: mov edi, dword ptr [rbp - 40] mov rax, qword ptr [rbp - 32] mov ecx, dword ptr [rbp - 16] sub ecx, 1 movsxd rcx, ecx mov esi, dword ptr [rax + 4*rcx] call min mov edx, eax add edx, 1 mov rax, qword ptr [rbp - 32] movsxd rcx, dword ptr [rbp - 16] mov dword ptr [rax + 4*rcx], edx .LBB0_7: mov eax, dword ptr [rbp - 40] mov dword ptr [rbp - 36], eax mov eax, dword ptr [rbp - 16] add eax, 1 mov dword ptr [rbp - 16], eax jmp .LBB0_3 .LBB0_9: jmp .LBB0_10 .LBB0_10: mov eax, dword ptr [rbp - 12] add eax, -1 mov dword ptr [rbp - 12], eax jmp .LBB0_1 .LBB0_11: mov rax, qword ptr [rbp - 32] mov ecx, dword ptr [rbp - 20] sub ecx, 1 movsxd rcx, ecx mov eax, dword ptr [rax + 4*rcx] add rsp, 48 pop rbp ret min: push rbp mov rbp, rsp mov dword ptr [rbp - 4], edi mov dword ptr [rbp - 8], esi mov eax, dword ptr [rbp - 4] cmp eax, dword ptr [rbp - 8] jge .LBB1_2 mov eax, dword ptr [rbp - 4] mov dword ptr [rbp - 12], eax jmp .LBB1_3 .LBB1_2: mov eax, dword ptr [rbp - 8] mov dword ptr [rbp - 12], eax .LBB1_3: mov eax, dword ptr [rbp - 12] pop rbp ret main: push rbp mov rbp, rsp sub rsp, 16 mov dword ptr [rbp - 4], 0 mov dword ptr [rbp - 8], edi mov qword ptr [rbp - 16], rsi cmp dword ptr [rbp - 8], 2 je .LBB2_2 mov rax, qword ptr [rip + stderr@GOTPCREL] mov rdi, qword ptr [rax] lea rsi, [rip + .L.str] mov al, 0 call fprintf@PLT mov edi, 4294967295 call exit@PLT .LBB2_2: mov rax, qword ptr [rbp - 16] mov rdi, qword ptr [rax + 8] call minInsertions mov esi, eax lea rdi, [rip + .L.str.1] mov al, 0 call printf@PLT xor eax, eax add rsp, 16 pop rbp ret .L.str: .asciz "Usage: ./test s\n" .L.str.1: .asciz "%d\n"
14,105
1,312
Minimum Insertion Steps to Make a String Palindrome
Hard
#include <stdio.h> #include <stdlib.h> #include <string.h> static inline int min(int a, int b) { return a < b ? a : b; } int minInsertions(char * s){ int i, j, len = strlen(s); int *dp = malloc(len * sizeof(int)); memset(dp, 0, len * sizeof(int)); for (i = len - 2; i >= 0; i--) { int left_down = 0; for (j = i + 1; j < len; j++) { int down = dp[j]; if (s[i] == s[j]) { dp[j] = left_down; } else { dp[j] = min(down, dp[j - 1]) + 1; } left_down = down; } } return dp[len - 1]; } int main(int argc, char **argv) { if (argc != 2) { fprintf(stderr, "Usage: ./test s\n"); exit(-1); } printf("%d\n", minInsertions(argv[1])); return 0; }
x86-64
-O1
x86-64 clang 21.1.0
minInsertions: push r15 push r14 push r13 push r12 push rbx mov rbx, rdi call strlen@PLT mov r14, rax mov r13, rax shl r13, 32 mov r12, r13 sar r12, 30 mov rdi, r12 call malloc@PLT mov r15, rax mov rdi, rax xor esi, esi mov rdx, r12 call memset@PLT cmp r14d, 2 jl .LBB0_8 mov eax, 4294967294 add eax, r14d lea ecx, [r14 - 1] mov edx, r14d and edx, 2147483647 dec rdx jmp .LBB0_2 .LBB0_7: dec rcx test rax, rax lea rax, [rax - 1] jle .LBB0_8 .LBB0_2: cmp rax, rdx jge .LBB0_7 movzx esi, byte ptr [rbx + rax] xor edi, edi mov r11, rcx mov r9, rax jmp .LBB0_4 .LBB0_6: mov dword ptr [r15 + 4*r8], r10d lea r11, [r8 + 1] mov r9, r8 cmp r11d, r14d jge .LBB0_7 .LBB0_4: mov r10d, edi mov r8, r11 mov edi, dword ptr [r15 + 4*r11] cmp sil, byte ptr [rbx + r11] je .LBB0_6 mov r9d, r9d mov r10d, dword ptr [r15 + 4*r9] cmp edi, r10d cmovl r10d, edi inc r10d jmp .LBB0_6 .LBB0_8: movabs rax, -4294967296 add r13, rax sar r13, 30 mov eax, dword ptr [r15 + r13] pop rbx pop r12 pop r13 pop r14 pop r15 ret main: push r15 push r14 push r13 push r12 push rbx cmp edi, 2 jne .LBB1_10 mov rbx, qword ptr [rsi + 8] mov rdi, rbx call strlen@PLT mov r14, rax mov r13, rax shl r13, 32 mov r12, r13 sar r12, 30 mov rdi, r12 call malloc@PLT mov r15, rax mov rdi, rax xor esi, esi mov rdx, r12 call memset@PLT cmp r14d, 2 jl .LBB1_9 mov eax, 4294967294 add eax, r14d lea ecx, [r14 - 1] mov edx, r14d and edx, 2147483647 dec rdx jmp .LBB1_3 .LBB1_8: dec rcx test rax, rax lea rax, [rax - 1] jle .LBB1_9 .LBB1_3: cmp rax, rdx jge .LBB1_8 movzx esi, byte ptr [rbx + rax] xor edi, edi mov r11, rcx mov r9, rax jmp .LBB1_5 .LBB1_7: mov dword ptr [r15 + 4*r8], r10d lea r11, [r8 + 1] mov r9, r8 cmp r11d, r14d jge .LBB1_8 .LBB1_5: mov r10d, edi mov r8, r11 mov edi, dword ptr [r15 + 4*r11] cmp sil, byte ptr [rbx + r11] je .LBB1_7 mov r9d, r9d mov r10d, dword ptr [r15 + 4*r9] cmp edi, r10d cmovl r10d, edi inc r10d jmp .LBB1_7 .LBB1_9: movabs rax, -4294967296 add r13, rax sar r13, 30 mov esi, dword ptr [r15 + r13] lea rdi, [rip + .L.str.1] xor eax, eax call printf@PLT xor eax, eax pop rbx pop r12 pop r13 pop r14 pop r15 ret .LBB1_10: mov rax, qword ptr [rip + stderr@GOTPCREL] mov rcx, qword ptr [rax] lea rdi, [rip + .L.str] mov esi, 16 mov edx, 1 call fwrite@PLT mov edi, -1 call exit@PLT .L.str: .asciz "Usage: ./test s\n" .L.str.1: .asciz "%d\n"
14,106
1,312
Minimum Insertion Steps to Make a String Palindrome
Hard
#include <stdio.h> #include <stdlib.h> #include <string.h> static inline int min(int a, int b) { return a < b ? a : b; } int minInsertions(char * s){ int i, j, len = strlen(s); int *dp = malloc(len * sizeof(int)); memset(dp, 0, len * sizeof(int)); for (i = len - 2; i >= 0; i--) { int left_down = 0; for (j = i + 1; j < len; j++) { int down = dp[j]; if (s[i] == s[j]) { dp[j] = left_down; } else { dp[j] = min(down, dp[j - 1]) + 1; } left_down = down; } } return dp[len - 1]; } int main(int argc, char **argv) { if (argc != 2) { fprintf(stderr, "Usage: ./test s\n"); exit(-1); } printf("%d\n", minInsertions(argv[1])); return 0; }
x86-64
-O2
x86-64 clang 21.1.0
minInsertions: push r15 push r14 push r12 push rbx push rax mov rbx, rdi call strlen@PLT mov r14, rax mov r15, rax shl r15, 32 mov rsi, r15 sar rsi, 30 mov edi, 1 call calloc@PLT cmp r14d, 2 jl .LBB0_8 mov ecx, 4294967294 add ecx, r14d lea edx, [r14 - 1] mov esi, r14d and esi, 2147483647 dec rsi jmp .LBB0_2 .LBB0_7: dec rdx test rcx, rcx lea rcx, [rcx - 1] jle .LBB0_8 .LBB0_2: cmp rcx, rsi jge .LBB0_7 movzx edi, byte ptr [rbx + rcx] xor r8d, r8d mov r12, rdx mov r10, rcx jmp .LBB0_4 .LBB0_6: mov dword ptr [rax + 4*r9], r11d lea r12, [r9 + 1] mov r10, r9 cmp r12d, r14d jge .LBB0_7 .LBB0_4: mov r11d, r8d mov r9, r12 mov r8d, dword ptr [rax + 4*r12] cmp dil, byte ptr [rbx + r12] je .LBB0_6 mov r10d, r10d mov r11d, dword ptr [rax + 4*r10] cmp r8d, r11d cmovl r11d, r8d inc r11d jmp .LBB0_6 .LBB0_8: movabs rcx, -4294967296 add r15, rcx sar r15, 30 mov eax, dword ptr [rax + r15] add rsp, 8 pop rbx pop r12 pop r14 pop r15 ret main: push r15 push r14 push r12 push rbx push rax cmp edi, 2 jne .LBB1_10 mov rbx, qword ptr [rsi + 8] mov rdi, rbx call strlen@PLT mov r14, rax mov r15, rax shl r15, 32 mov rsi, r15 sar rsi, 30 mov edi, 1 call calloc@PLT cmp r14d, 2 jl .LBB1_9 mov ecx, 4294967294 add ecx, r14d lea edx, [r14 - 1] mov esi, r14d and esi, 2147483647 dec rsi jmp .LBB1_3 .LBB1_8: dec rdx test rcx, rcx lea rcx, [rcx - 1] jle .LBB1_9 .LBB1_3: cmp rcx, rsi jge .LBB1_8 movzx edi, byte ptr [rbx + rcx] xor r8d, r8d mov r12, rdx mov r10, rcx jmp .LBB1_5 .LBB1_7: mov dword ptr [rax + 4*r9], r11d lea r12, [r9 + 1] mov r10, r9 cmp r12d, r14d jge .LBB1_8 .LBB1_5: mov r11d, r8d mov r9, r12 mov r8d, dword ptr [rax + 4*r12] cmp dil, byte ptr [rbx + r12] je .LBB1_7 mov r10d, r10d mov r11d, dword ptr [rax + 4*r10] cmp r8d, r11d cmovl r11d, r8d inc r11d jmp .LBB1_7 .LBB1_9: movabs rcx, -4294967296 add r15, rcx sar r15, 30 mov esi, dword ptr [rax + r15] lea rdi, [rip + .L.str.1] xor eax, eax call printf@PLT xor eax, eax add rsp, 8 pop rbx pop r12 pop r14 pop r15 ret .LBB1_10: mov rax, qword ptr [rip + stderr@GOTPCREL] mov rcx, qword ptr [rax] lea rdi, [rip + .L.str] mov esi, 16 mov edx, 1 call fwrite@PLT mov edi, -1 call exit@PLT .L.str: .asciz "Usage: ./test s\n" .L.str.1: .asciz "%d\n"
14,107
1,312
Minimum Insertion Steps to Make a String Palindrome
Hard
#include <stdio.h> #include <stdlib.h> #include <string.h> static inline int min(int a, int b) { return a < b ? a : b; } int minInsertions(char * s){ int i, j, len = strlen(s); int *dp = malloc(len * sizeof(int)); memset(dp, 0, len * sizeof(int)); for (i = len - 2; i >= 0; i--) { int left_down = 0; for (j = i + 1; j < len; j++) { int down = dp[j]; if (s[i] == s[j]) { dp[j] = left_down; } else { dp[j] = min(down, dp[j - 1]) + 1; } left_down = down; } } return dp[len - 1]; } int main(int argc, char **argv) { if (argc != 2) { fprintf(stderr, "Usage: ./test s\n"); exit(-1); } printf("%d\n", minInsertions(argv[1])); return 0; }
x86-64
-O3
x86-64 clang 21.1.0
minInsertions: push r15 push r14 push r12 push rbx push rax mov rbx, rdi call strlen@PLT mov r14, rax mov r15, rax shl r15, 32 mov rsi, r15 sar rsi, 30 mov edi, 1 call calloc@PLT cmp r14d, 2 jl .LBB0_8 mov ecx, 4294967294 add ecx, r14d lea edx, [r14 - 1] mov esi, r14d and esi, 2147483647 dec rsi jmp .LBB0_2 .LBB0_7: dec rdx test rcx, rcx lea rcx, [rcx - 1] jle .LBB0_8 .LBB0_2: cmp rcx, rsi jge .LBB0_7 movzx edi, byte ptr [rbx + rcx] xor r8d, r8d mov r12, rdx mov r10, rcx jmp .LBB0_4 .LBB0_6: mov dword ptr [rax + 4*r9], r11d lea r12, [r9 + 1] mov r10, r9 cmp r12d, r14d jge .LBB0_7 .LBB0_4: mov r11d, r8d mov r9, r12 mov r8d, dword ptr [rax + 4*r12] cmp dil, byte ptr [rbx + r12] je .LBB0_6 mov r10d, r10d mov r11d, dword ptr [rax + 4*r10] cmp r8d, r11d cmovl r11d, r8d inc r11d jmp .LBB0_6 .LBB0_8: movabs rcx, -4294967296 add r15, rcx sar r15, 30 mov eax, dword ptr [rax + r15] add rsp, 8 pop rbx pop r12 pop r14 pop r15 ret main: push r15 push r14 push r12 push rbx push rax cmp edi, 2 jne .LBB1_10 mov rbx, qword ptr [rsi + 8] mov rdi, rbx call strlen@PLT mov r14, rax mov r15, rax shl r15, 32 mov rsi, r15 sar rsi, 30 mov edi, 1 call calloc@PLT cmp r14d, 2 jl .LBB1_9 mov ecx, 4294967294 add ecx, r14d lea edx, [r14 - 1] mov esi, r14d and esi, 2147483647 dec rsi jmp .LBB1_3 .LBB1_8: dec rdx test rcx, rcx lea rcx, [rcx - 1] jle .LBB1_9 .LBB1_3: cmp rcx, rsi jge .LBB1_8 movzx edi, byte ptr [rbx + rcx] xor r8d, r8d mov r12, rdx mov r10, rcx jmp .LBB1_5 .LBB1_7: mov dword ptr [rax + 4*r9], r11d lea r12, [r9 + 1] mov r10, r9 cmp r12d, r14d jge .LBB1_8 .LBB1_5: mov r11d, r8d mov r9, r12 mov r8d, dword ptr [rax + 4*r12] cmp dil, byte ptr [rbx + r12] je .LBB1_7 mov r10d, r10d mov r11d, dword ptr [rax + 4*r10] cmp r8d, r11d cmovl r11d, r8d inc r11d jmp .LBB1_7 .LBB1_9: movabs rcx, -4294967296 add r15, rcx sar r15, 30 mov esi, dword ptr [rax + r15] lea rdi, [rip + .L.str.1] xor eax, eax call printf@PLT xor eax, eax add rsp, 8 pop rbx pop r12 pop r14 pop r15 ret .LBB1_10: mov rax, qword ptr [rip + stderr@GOTPCREL] mov rcx, qword ptr [rax] lea rdi, [rip + .L.str] mov esi, 16 mov edx, 1 call fwrite@PLT mov edi, -1 call exit@PLT .L.str: .asciz "Usage: ./test s\n" .L.str.1: .asciz "%d\n"
14,108
1,312
Minimum Insertion Steps to Make a String Palindrome
Hard
#include <stdio.h> #include <stdlib.h> #include <string.h> static inline int min(int a, int b) { return a < b ? a : b; } int minInsertions(char * s){ int i, j, len = strlen(s); int *dp = malloc(len * sizeof(int)); memset(dp, 0, len * sizeof(int)); for (i = len - 2; i >= 0; i--) { int left_down = 0; for (j = i + 1; j < len; j++) { int down = dp[j]; if (s[i] == s[j]) { dp[j] = left_down; } else { dp[j] = min(down, dp[j - 1]) + 1; } left_down = down; } } return dp[len - 1]; } int main(int argc, char **argv) { if (argc != 2) { fprintf(stderr, "Usage: ./test s\n"); exit(-1); } printf("%d\n", minInsertions(argv[1])); return 0; }
x86-64
-O0
x86-64 gcc 15.2
min: push rbp mov rbp, rsp mov DWORD PTR [rbp-4], edi mov DWORD PTR [rbp-8], esi mov edx, DWORD PTR [rbp-8] mov eax, DWORD PTR [rbp-4] cmp edx, eax cmovle eax, edx pop rbp ret minInsertions: push rbp mov rbp, rsp sub rsp, 48 mov QWORD PTR [rbp-40], rdi mov rax, QWORD PTR [rbp-40] mov rdi, rax call strlen mov DWORD PTR [rbp-16], eax mov eax, DWORD PTR [rbp-16] cdqe sal rax, 2 mov rdi, rax call malloc mov QWORD PTR [rbp-24], rax mov eax, DWORD PTR [rbp-16] cdqe lea rdx, [0+rax*4] mov rax, QWORD PTR [rbp-24] mov esi, 0 mov rdi, rax call memset mov eax, DWORD PTR [rbp-16] sub eax, 2 mov DWORD PTR [rbp-4], eax jmp .L4 .L9: mov DWORD PTR [rbp-12], 0 mov eax, DWORD PTR [rbp-4] add eax, 1 mov DWORD PTR [rbp-8], eax jmp .L5 .L8: mov eax, DWORD PTR [rbp-8] cdqe lea rdx, [0+rax*4] mov rax, QWORD PTR [rbp-24] add rax, rdx mov eax, DWORD PTR [rax] mov DWORD PTR [rbp-28], eax mov eax, DWORD PTR [rbp-4] movsx rdx, eax mov rax, QWORD PTR [rbp-40] add rax, rdx movzx edx, BYTE PTR [rax] mov eax, DWORD PTR [rbp-8] movsx rcx, eax mov rax, QWORD PTR [rbp-40] add rax, rcx movzx eax, BYTE PTR [rax] cmp dl, al jne .L6 mov eax, DWORD PTR [rbp-8] cdqe lea rdx, [0+rax*4] mov rax, QWORD PTR [rbp-24] add rdx, rax mov eax, DWORD PTR [rbp-12] mov DWORD PTR [rdx], eax jmp .L7 .L6: mov eax, DWORD PTR [rbp-8] cdqe sal rax, 2 lea rdx, [rax-4] mov rax, QWORD PTR [rbp-24] add rax, rdx mov edx, DWORD PTR [rax] mov eax, DWORD PTR [rbp-28] mov esi, edx mov edi, eax call min mov edx, DWORD PTR [rbp-8] movsx rdx, edx lea rcx, [0+rdx*4] mov rdx, QWORD PTR [rbp-24] add rdx, rcx add eax, 1 mov DWORD PTR [rdx], eax .L7: mov eax, DWORD PTR [rbp-28] mov DWORD PTR [rbp-12], eax add DWORD PTR [rbp-8], 1 .L5: mov eax, DWORD PTR [rbp-8] cmp eax, DWORD PTR [rbp-16] jl .L8 sub DWORD PTR [rbp-4], 1 .L4: cmp DWORD PTR [rbp-4], 0 jns .L9 mov eax, DWORD PTR [rbp-16] cdqe sal rax, 2 lea rdx, [rax-4] mov rax, QWORD PTR [rbp-24] add rax, rdx mov eax, DWORD PTR [rax] leave ret .LC0: .string "Usage: ./test s\n" .LC1: .string "%d\n" main: push rbp mov rbp, rsp sub rsp, 16 mov DWORD PTR [rbp-4], edi mov QWORD PTR [rbp-16], rsi cmp DWORD PTR [rbp-4], 2 je .L12 mov rax, QWORD PTR stderr[rip] mov rcx, rax mov edx, 16 mov esi, 1 mov edi, OFFSET FLAT:.LC0 call fwrite mov edi, -1 call exit .L12: mov rax, QWORD PTR [rbp-16] add rax, 8 mov rax, QWORD PTR [rax] mov rdi, rax call minInsertions mov esi, eax mov edi, OFFSET FLAT:.LC1 mov eax, 0 call printf mov eax, 0 leave ret
14,109
1,312
Minimum Insertion Steps to Make a String Palindrome
Hard
#include <stdio.h> #include <stdlib.h> #include <string.h> static inline int min(int a, int b) { return a < b ? a : b; } int minInsertions(char * s){ int i, j, len = strlen(s); int *dp = malloc(len * sizeof(int)); memset(dp, 0, len * sizeof(int)); for (i = len - 2; i >= 0; i--) { int left_down = 0; for (j = i + 1; j < len; j++) { int down = dp[j]; if (s[i] == s[j]) { dp[j] = left_down; } else { dp[j] = min(down, dp[j - 1]) + 1; } left_down = down; } } return dp[len - 1]; } int main(int argc, char **argv) { if (argc != 2) { fprintf(stderr, "Usage: ./test s\n"); exit(-1); } printf("%d\n", minInsertions(argv[1])); return 0; }
x86-64
-O1
x86-64 gcc 15.2
minInsertions: push r15 push r14 push r13 push r12 push rbp push rbx sub rsp, 24 mov rbx, rdi call strlen mov r13, rax mov ebp, eax movsx r12, eax lea r14, [0+r12*4] mov rdi, r14 call malloc mov r15, rax mov QWORD PTR [rsp], rax mov rdx, r14 mov esi, 0 mov rdi, rax call memset cmp r13d, 1 jle .L2 sub r13d, 1 lea r11, [r15-8+r14] lea r10, [rbx-1+r12] mov edi, r13d lea r14, [r15+4] mov r15d, 0 mov QWORD PTR [rsp+8], r12 jmp .L6 .L13: mov eax, DWORD PTR [rdx] cmp ecx, eax cmovle eax, ecx add eax, 1 .L4: mov DWORD PTR [r8+4], eax add rdx, 4 add rsi, 1 cmp rdx, r9 je .L12 .L5: mov r8, rdx mov eax, ecx mov ecx, DWORD PTR [rdx+4] cmp r12b, BYTE PTR [rsi] jne .L13 jmp .L4 .L12: sub edi, 1 je .L10 sub r11, 4 sub r10, 1 .L6: cmp ebp, edi jle .L3 movzx r12d, BYTE PTR [r10-1] mov rdx, r10 sub rdx, rbx mov eax, r13d sub eax, edi lea rax, [rdx-1+rax] lea r9, [r14+rax*4] mov rsi, r10 mov rdx, r11 mov ecx, r15d jmp .L5 .L10: mov r12, QWORD PTR [rsp+8] .L2: mov rax, QWORD PTR [rsp] mov eax, DWORD PTR [rax-4+r12*4] add rsp, 24 pop rbx pop rbp pop r12 pop r13 pop r14 pop r15 ret .L3: sub edi, 1 sub r11, 4 sub r10, 1 jmp .L6 .LC0: .string "Usage: ./test s\n" .LC1: .string "%d\n" main: sub rsp, 8 cmp edi, 2 jne .L17 mov rdi, QWORD PTR [rsi+8] call minInsertions mov esi, eax mov edi, OFFSET FLAT:.LC1 mov eax, 0 call printf mov eax, 0 add rsp, 8 ret .L17: mov rcx, QWORD PTR stderr[rip] mov edx, 16 mov esi, 1 mov edi, OFFSET FLAT:.LC0 call fwrite mov edi, -1 call exit
14,110
1,312
Minimum Insertion Steps to Make a String Palindrome
Hard
#include <stdio.h> #include <stdlib.h> #include <string.h> static inline int min(int a, int b) { return a < b ? a : b; } int minInsertions(char * s){ int i, j, len = strlen(s); int *dp = malloc(len * sizeof(int)); memset(dp, 0, len * sizeof(int)); for (i = len - 2; i >= 0; i--) { int left_down = 0; for (j = i + 1; j < len; j++) { int down = dp[j]; if (s[i] == s[j]) { dp[j] = left_down; } else { dp[j] = min(down, dp[j - 1]) + 1; } left_down = down; } } return dp[len - 1]; } int main(int argc, char **argv) { if (argc != 2) { fprintf(stderr, "Usage: ./test s\n"); exit(-1); } printf("%d\n", minInsertions(argv[1])); return 0; }
x86-64
-O2
x86-64 gcc 15.2
minInsertions: push r15 push r14 push r13 push r12 push rbp mov rbp, rdi push rbx sub rsp, 8 call strlen mov esi, 1 movsx r12, eax mov rbx, rax lea rdi, [0+r12*4] call calloc mov r15, rax cmp ebx, 1 jle .L2 lea r9d, [rbx-1] lea r11, [rax-8+r12*4] lea r10, [rbp-1+r12] lea r14, [rax+4] cmp ebx, r9d jle .L3 .L16: mov rdx, r10 lea eax, [rbx-1] movzx r8d, BYTE PTR [r10-1] mov rsi, r10 sub rdx, rbp sub eax, r9d xor ecx, ecx lea rax, [rdx-1+rax] mov rdx, r11 lea rdi, [r14+rax*4] .L5: mov eax, ecx mov ecx, DWORD PTR [rdx+4] cmp r8b, BYTE PTR [rsi] je .L4 mov eax, DWORD PTR [rdx] cmp ecx, eax cmovle eax, ecx add eax, 1 .L4: mov DWORD PTR [rdx+4], eax add rdx, 4 add rsi, 1 cmp rdx, rdi jne .L5 sub r9d, 1 je .L2 .L14: sub r11, 4 sub r10, 1 cmp ebx, r9d jg .L16 .L3: sub r9d, 1 jmp .L14 .L2: mov eax, DWORD PTR [r15-4+r12*4] add rsp, 8 pop rbx pop rbp pop r12 pop r13 pop r14 pop r15 ret .LC0: .string "Usage: ./test s\n" .LC1: .string "%d\n" main: sub rsp, 8 cmp edi, 2 jne .L20 mov rdi, QWORD PTR [rsi+8] call minInsertions mov edi, OFFSET FLAT:.LC1 mov esi, eax xor eax, eax call printf xor eax, eax add rsp, 8 ret .L20: mov edi, OFFSET FLAT:.LC0 mov edx, 16 mov esi, 1 mov rcx, QWORD PTR stderr[rip] call fwrite mov edi, -1 call exit
14,111
1,312
Minimum Insertion Steps to Make a String Palindrome
Hard
#include <stdio.h> #include <stdlib.h> #include <string.h> static inline int min(int a, int b) { return a < b ? a : b; } int minInsertions(char * s){ int i, j, len = strlen(s); int *dp = malloc(len * sizeof(int)); memset(dp, 0, len * sizeof(int)); for (i = len - 2; i >= 0; i--) { int left_down = 0; for (j = i + 1; j < len; j++) { int down = dp[j]; if (s[i] == s[j]) { dp[j] = left_down; } else { dp[j] = min(down, dp[j - 1]) + 1; } left_down = down; } } return dp[len - 1]; } int main(int argc, char **argv) { if (argc != 2) { fprintf(stderr, "Usage: ./test s\n"); exit(-1); } printf("%d\n", minInsertions(argv[1])); return 0; }
x86-64
-O3
x86-64 gcc 15.2
minInsertions: push r15 push r14 push r13 push r12 push rbp mov rbp, rdi push rbx sub rsp, 8 call strlen mov esi, 1 movsx r12, eax mov rbx, rax lea rdi, [0+r12*4] call calloc mov r15, rax cmp ebx, 1 jle .L2 lea r9d, [rbx-1] lea r11, [rax-8+r12*4] lea r10, [rbp-1+r12] lea r14, [rax+4] cmp ebx, r9d jle .L3 .L19: mov rdx, r10 lea eax, [rbx-1] movzx edi, BYTE PTR [r10-1] mov rsi, r10 sub rdx, rbp sub eax, r9d lea rax, [rdx-1+rax] xor edx, edx lea r8, [r14+rax*4] mov rax, r11 jmp .L7 .L18: mov edx, DWORD PTR [rax] cmp ecx, edx cmovle edx, ecx add rax, 4 add edx, 1 mov DWORD PTR [rax], edx cmp r8, rax je .L5 .L16: add rsi, 1 mov edx, ecx .L7: mov ecx, DWORD PTR [rax+4] cmp dil, BYTE PTR [rsi] jne .L18 mov DWORD PTR [rax+4], edx add rax, 4 cmp r8, rax jne .L16 .L5: sub r9d, 1 je .L2 .L15: sub r11, 4 sub r10, 1 cmp ebx, r9d jg .L19 .L3: sub r9d, 1 jmp .L15 .L2: mov eax, DWORD PTR [r15-4+r12*4] add rsp, 8 pop rbx pop rbp pop r12 pop r13 pop r14 pop r15 ret .LC0: .string "Usage: ./test s\n" .LC1: .string "%d\n" main: sub rsp, 8 cmp edi, 2 jne .L23 mov rdi, QWORD PTR [rsi+8] call minInsertions mov edi, OFFSET FLAT:.LC1 mov esi, eax xor eax, eax call printf xor eax, eax add rsp, 8 ret .L23: mov edi, OFFSET FLAT:.LC0 mov edx, 16 mov esi, 1 mov rcx, QWORD PTR stderr[rip] call fwrite mov edi, -1 call exit