Maximum Swap

Maximum Swap - Problem

Math Greedy Medium

You are given an integer num. You can swap two digits at most once to get the maximum valued number.

Return the maximum valued number you can get.

Input & Output

Example 1 — Basic Swap

$ Input: num = 2736

› Output: 7236

💡 Note: Swap the first digit 2 with the second digit 7 to get 7236, which is the maximum possible value with a single swap.

Example 2 — No Improvement Possible

$ Input: num = 9973

› Output: 9973

💡 Note: The digits are already in optimal order from left to right, no single swap can improve the value.

Example 3 — Single Digit

$ Input: num = 1

› Output: 1

💡 Note: Single digit number cannot be improved by swapping.

Constraints

0 ≤ num ≤ 10⁸

Visualization

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Asked in

F Facebook 25 G Google 15

The key insight is to use a greedy approach: record the rightmost position of each digit, then scan from left to find the first position that can be improved by swapping with a larger digit from the right. Best approach is greedy single-pass. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Try All Swaps

⏱️ Time: O(n²) Space: O(n)

Convert the number to a string, try swapping every pair of positions, convert back to integer and track the maximum value found. This guarantees finding the optimal solution by checking all possibilities.

Greedy - Find Best Single Swap

⏱️ Time: O(n) Space: O(1)

Track the rightmost position of each digit, then scan from left to find the first position where we can improve by swapping with a larger digit from the right. This ensures we make the most impactful swap.

Brute Force - Try All Swaps — Algorithm Steps

Step 1: Convert number to string for easy digit manipulation
Step 2: Try swapping every pair of positions (i,j) where i < j
Step 3: Convert each result back to integer and track maximum
Step 4: Return the maximum value found (or original if no improvement)

Visualization

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Step-by-Step Walkthrough

Original

Start with the given number

Try Swaps

Test every pair of positions

Track Maximum

Keep the best result found

Code -

solution.c — C

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

int solution(int num) {
    char digits[20];
    sprintf(digits, "%d", num);
    int n = strlen(digits);
    int maxNum = num;
    
    // Try all possible swaps
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            // Swap digits at positions i and j
            char temp = digits[i];
            digits[i] = digits[j];
            digits[j] = temp;
            // Convert back to integer
            int currentNum = atoi(digits);
            if (currentNum > maxNum) {
                maxNum = currentNum;
            }
            // Swap back
            temp = digits[i];
            digits[i] = digits[j];
            digits[j] = temp;
        }
    }
    
    return maxNum;
}

int main() {
    int num;
    scanf("%d", &num);
    int result = solution(num);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops to try all pairs of positions - n*(n-1)/2 swaps

⚠ Quadratic Growth

Space Complexity

O(n)

String conversion and temporary arrays for each swap attempt

⚡ Linearithmic Space

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Medium Frequency

~15 min Avg. Time

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Try All Swaps — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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