Minimum Possible Integer After at Most K Adjacent Swaps On Digits

Minimum Possible Integer After at Most K Adjacent Swaps On Digits - Problem

You are given a string num representing the digits of a very large integer and an integer k. You are allowed to swap any two adjacent digits of the integer at most k times.

Return the minimum integer you can obtain also as a string.

Input & Output

Example 1 — Basic Greedy Selection

$ Input: num = "4321", k = 4

› Output: "1342"

💡 Note: Move 1 to front (3 swaps: 4321→4312→4132→1432→1342), then move 2 forward (1 swap: 1342→1324 would exceed k), so result is "1342"

Example 2 — Limited Swaps

$ Input: num = "100", k = 1

› Output: "010"

💡 Note: Can only make 1 swap. Best option is to swap first two digits: 100→010

Example 3 — No Swaps Needed

$ Input: num = "1234", k = 3

› Output: "1234"

💡 Note: Already in minimum order, no swaps needed

Constraints

1 ≤ num.length ≤ 3 × 10⁴
num consists only of digits
0 ≤ k ≤ 10⁹

Visualization

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Asked in

G Google 15 f Facebook 8 a Amazon 6

The key insight is to greedily move the smallest available digits to the front first. The optimal approach uses a Fenwick Tree to efficiently track position costs during swaps. Time: O(n log n), Space: O(n)

Common Approaches

✓ Brute Force with DFS

⏱️ Time: O(k! × n) Space: O(k × n)

For each position, try all possible adjacent swaps and recursively find the minimum result. This explores all possible ways to use the k swaps.

Greedy Approach

⏱️ Time: O(n²) Space: O(1)

For each position from left to right, find the smallest digit within reach (considering remaining swaps) and bubble it to the current position using adjacent swaps.

Optimized with Fenwick Tree

⏱️ Time: O(n log n) Space: O(n)

Use a greedy approach combined with Fenwick Tree (Binary Indexed Tree) to efficiently track and update positions when moving digits. This reduces the time complexity for position queries.

Brute Force with DFS — Algorithm Steps

Step 1: For current state, try all adjacent swaps
Step 2: Recursively solve with remaining swaps
Step 3: Return the lexicographically smallest result

Visualization

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Step-by-Step Walkthrough

Initial State

Start with original string and k swaps

Try All Swaps

For each adjacent pair, try swapping

Recurse

Continue with remaining swaps and find minimum

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char* dfs(char* s, int swapsLeft, char* result) {
    if (swapsLeft == 0) {
        if (strcmp(s, result) < 0) {
            strcpy(result, s);
        }
        return result;
    }
    
    int len = strlen(s);
    // Try all adjacent swaps
    for (int i = 0; i < len - 1; i++) {
        // Swap adjacent digits
        char temp = s[i];
        s[i] = s[i + 1];
        s[i + 1] = temp;
        
        // Recursively solve with one less swap
        dfs(s, swapsLeft - 1, result);
        
        // Backtrack
        temp = s[i];
        s[i] = s[i + 1];
        s[i + 1] = temp;
    }
    
    return result;
}

char* solution(char* num, int k) {
    char* result = malloc(strlen(num) + 1);
    strcpy(result, num);
    char* temp = malloc(strlen(num) + 1);
    strcpy(temp, num);
    dfs(temp, k, result);
    free(temp);
    return result;
}

int main() {
    char num[100001];
    int k;
    fgets(num, sizeof(num), stdin);
    num[strcspn(num, "\n")] = 0;
    scanf("%d", &k);
    
    char* result = solution(num, k);
    printf("%s\n", result);
    free(result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(k! × n)

Each swap creates new branches, leading to factorial combinations

⚠ Quadratic Growth

Space Complexity

O(k × n)

Recursion stack depth up to k levels, each storing string of length n

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force with DFS — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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