Shuffle String - Practice Coding Problems

Shuffle String - Problem

Array String Easy

You are given a string s and an integer array indices of the same length. The string s will be shuffled such that the character at the ith position moves to indices[i] in the shuffled string.

Return the shuffled string.

Input & Output

Example 1 — Basic Shuffling

$ Input: s = "codeleet", indices = [4,5,6,7,0,2,1,3]

› Output: "leetcode"

💡 Note: Characters are rearranged: s[0]='c' goes to indices[0]=4, s[1]='o' goes to indices[1]=5, etc. Result: "leetcode"

Example 2 — Short String

$ Input: s = "abc", indices = [0,1,2]

› Output: "abc"

💡 Note: Each character stays in the same position: 'a' at 0, 'b' at 1, 'c' at 2

Example 3 — Complete Reversal

$ Input: s = "aiohn", indices = [3,1,4,2,0]

› Output: "nihao"

💡 Note: s[0]='a' → pos 3, s[1]='i' → pos 1, s[2]='o' → pos 4, s[3]='h' → pos 2, s[4]='n' → pos 0

Constraints

s.length == indices.length
1 ≤ s.length ≤ 100
s contains only lowercase English letters
0 ≤ indices[i] < s.length
All values of indices are unique

Visualization

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Understanding the Visualization

Input

String 'codeleet' and indices [4,5,6,7,0,2,1,3]

Mapping

Each character moves to its target position

Output

Shuffled string 'leetcode'

Key Takeaway

🎯 Key Insight: Each character has a predetermined target position - just place them directly without complex rearrangement logic

Asked in

M Microsoft 15 a Amazon 12 G Google 8

The key insight is that each character has a predetermined destination position. The optimal approach creates a result array and places each character directly at its target position in O(n) time. Best approach: Character by Character Placement - Time: O(n), Space: O(n)

Common Approaches

Approach	Time	Space	Notes
✓ Direct String Building	O(n²)	O(1)	Build result string directly without intermediate array
Character by Character Placement	O(n)	O(n)	Place each character one by one at its target position

Direct String Building — Algorithm Steps

Iterate through target positions 0 to n-1
Find which original character goes to each position
Build result string position by position

Visualization

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Step-by-Step Walkthrough

Target Position 0

Find which original character maps to position 0

Target Position 1

Find which original character maps to position 1

Continue

Build string position by position

Code -

solution.c — C

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

char* solution(char* s, int* indices, int length) {
    char* result = (char*)malloc((length + 1) * sizeof(char));
    result[length] = '\0';
    
    for (int targetPos = 0; targetPos < length; targetPos++) {
        // Find which character should go to targetPos
        for (int i = 0; i < length; i++) {
            if (indices[i] == targetPos) {
                result[targetPos] = s[i];
                break;
            }
        }
    }
    
    return result;
}

int main() {
    char s[1001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    
    char line[10001];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int indices[1000];
    int length = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        indices[length++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    char* result = solution(s, indices, length);
    printf("%s\n", result);
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each position, we search through indices array to find the source

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables for iteration, result string not counted as extra space

✓ Linear Space

72.8K Views

Medium Frequency

~8 min Avg. Time

2.1K Likes

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Direct String Building — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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