Rearrange K Substrings to Form Target String

Rearrange K Substrings to Form Target String - Problem

You are given two strings s and t, both of which are anagrams of each other, and an integer k.

Your task is to determine whether it is possible to split the string s into k equal-sized substrings, rearrange the substrings, and concatenate them in any order to create a new string that matches the given string t.

Return true if this is possible, otherwise, return false.

Note: An anagram is a word or phrase formed by rearranging the letters of a different word or phrase, using all the original letters exactly once. A substring is a contiguous non-empty sequence of characters within a string.

Input & Output

Example 1 — Basic Rearrangement

$ Input: s = "abcd", t = "cdab", k = 2

› Output: true

💡 Note: Split s into ["ab", "cd"] and t into ["cd", "ab"] with k=2. We can rearrange s chunks ["ab", "cd"] to ["cd", "ab"] to form t.

Example 2 — Impossible Rearrangement

$ Input: s = "abcd", t = "acbd", k = 2

› Output: false

💡 Note: Split s into ["ab", "cd"] and t into ["ac", "bd"] with k=2. No rearrangement of s chunks can form t chunks.

Example 3 — Single Character Chunks

$ Input: s = "abc", t = "bca", k = 3

› Output: true

💡 Note: Split s into ["a", "b", "c"] and t into ["b", "c", "a"] with k=3. We can rearrange s chunks to match t chunks.

Constraints

1 ≤ s.length, t.length ≤ 1000
1 ≤ k ≤ s.length
s.length is divisible by k
s and t consist of lowercase English letters only
s and t are anagrams of each other

Visualization

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Understanding the Visualization

Input

Two anagram strings s and t, and integer k

Split

Divide both strings into k equal-length substrings

Check

Verify if substring multisets match

Key Takeaway

🎯 Key Insight: If k substrings from s can be rearranged to form t, both strings must have identical multisets of k-length substrings

Asked in

G Google 25 M Microsoft 20 a Amazon 15

The key insight is that if we can rearrange k substrings from s to form t, then both strings must have the same multiset of k-length substrings. Best approach is frequency counting - split both strings, count substring frequencies, and compare. Time: O(n + k log k), Space: O(k)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Generate All Permutations	O(k! × n)	O(n)	Generate all possible arrangements of k substrings from s and check if any matches t
Sort and Compare Optimization	O(n + k log k)	O(n)	Sort both sets of k substrings and compare them directly for equality
Frequency Counting Optimization	O(n + k log k)	O(n + k)	Compare frequency of k-length substrings in both s and t to determine if rearrangement is possible

Brute Force - Generate All Permutations — Algorithm Steps

Split s into k equal-length substrings
Generate all k! permutations of substrings
Check each permutation concatenation against t

Visualization

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Step-by-Step Walkthrough

Split s

Divide s into k equal substrings

Generate Permutations

Create all k! arrangements

Check Match

Test each arrangement against t

Code -

solution.c — C

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

bool checkPermutation(char chunks[][100], int k, int chunkSize, char* t, bool used[], char* current, int pos, int currentLen) {
    if (pos == k) {
        current[currentLen] = '\0';
        return strcmp(current, t) == 0;
    }
    
    for (int i = 0; i < k; i++) {
        if (!used[i]) {
            used[i] = true;
            strcat(current, chunks[i]);
            if (checkPermutation(chunks, k, chunkSize, t, used, current, pos + 1, currentLen + chunkSize)) {
                return true;
            }
            current[currentLen] = '\0';
            used[i] = false;
        }
    }
    return false;
}

bool solution(char* s, char* t, int k) {
    int n = strlen(s);
    if (n % k != 0) return false;
    
    int chunkSize = n / k;
    char chunks[k][100];
    bool used[k];
    char current[1000] = "";
    
    for (int i = 0; i < k; i++) {
        strncpy(chunks[i], s + i * chunkSize, chunkSize);
        chunks[i][chunkSize] = '\0';
        used[i] = false;
    }
    
    return checkPermutation(chunks, k, chunkSize, t, used, current, 0, 0);
}

int main() {
    char s[1000], t[1000];
    int k;
    
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    fgets(t, sizeof(t), stdin);
    t[strcspn(t, "\n")] = 0;
    scanf("%d", &k);
    
    bool result = solution(s, t, k);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(k! × n)

k! permutations, each requiring O(n) time to concatenate and compare

✓ Linear Growth

Space Complexity

O(n)

Store k substrings and temporary concatenation results

⚡ Linearithmic Space

28.4K Views

Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Generate All Permutations — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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