String Transformation - Practice Coding Problems

String Transformation - Problem

You are given two strings s and t of equal length n. You can perform the following operation on the string s:

Remove a suffix of s of length l where 0 < l < n and append it at the start of s.

For example, let s = 'abcd' then in one operation you can remove the suffix 'cd' and append it in front of s making s = 'cdab'.

You are also given an integer k. Return the number of ways in which s can be transformed into t in exactly k operations.

Since the answer can be large, return it modulo 10⁹ + 7.

Input & Output

Example 1 — Basic Transformation

$ Input: s = "abcd", t = "cdab", k = 1

› Output: 1

💡 Note: Remove suffix "cd" (length 2) and append to front: "abcd" → "cdab". There's exactly 1 way to transform s to t in 1 operation.

Example 2 — Multiple Operations

$ Input: s = "abc", t = "bca", k = 2

› Output: 1

💡 Note: One possible path: "abc" → remove "c" → "cab" → remove "ab" → "bca". Need to count all paths with exactly 2 operations.

Example 3 — Impossible Target

$ Input: s = "abc", t = "xyz", k = 1

› Output: 0

💡 Note: Target "xyz" is not a rotation of "abc", so it's impossible to reach regardless of operations.

Constraints

2 ≤ s.length ≤ 1000
t.length = s.length
1 ≤ k ≤ 1000
s and t consist of lowercase English letters

Visualization

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Understanding the Visualization

Input

Source string s, target string t, and exactly k operations

Operations

Remove suffix and append to front creates rotations

Count Ways

Find number of paths from s to t in exactly k steps

Key Takeaway

🎯 Key Insight: Each suffix operation creates a string rotation - use DP to count paths efficiently

Asked in

G Google 45 M Microsoft 32 a Amazon 28

The key insight is recognizing that each operation creates a rotation of the original string. The best approach uses dynamic programming with rotation indices to efficiently count transformation paths. Time: O(n² × k), Space: O(n × k).

Common Approaches

Approach	Time	Space	Notes
✓ Memoized Recursion	O(n^k × n)	O(n^k × n)	Cache results for (current_string, operations_left) pairs to avoid recomputation
Dynamic Programming with Rotation Index	O(n² × k)	O(n × k)	Use DP table where dp[i][j] represents ways to reach rotation i in j operations
Recursive Brute Force	O((n-1)^k)	O(k)	Try all possible suffix removals at each step recursively

Memoized Recursion — Algorithm Steps

Use memo dictionary with (string, operations_left) as key
Before recursive call, check if result already computed
Store and return computed results to avoid duplicate work

Visualization

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Step-by-Step Walkthrough

First Computation

Compute result for (string, operations_left) and store in cache

Cache Hit

When same state appears again, return cached result

Efficiency Gain

Avoid exponential recomputation of identical subproblems

Code -

solution.c — C

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

#define MOD 1000000007
#define MAX_STATES 100000

typedef struct {
    char state[1001];
    int ops;
    int result;
} MemoEntry;

MemoEntry memo[MAX_STATES];
int memoSize = 0;

int findMemo(char* current, int operationsLeft) {
    for (int i = 0; i < memoSize; i++) {
        if (strcmp(memo[i].state, current) == 0 && memo[i].ops == operationsLeft) {
            return memo[i].result;
        }
    }
    return -1;
}

void addMemo(char* current, int operationsLeft, int result) {
    if (memoSize < MAX_STATES) {
        strcpy(memo[memoSize].state, current);
        memo[memoSize].ops = operationsLeft;
        memo[memoSize].result = result;
        memoSize++;
    }
}

int dp(char* current, char* target, int operationsLeft, int n) {
    if (operationsLeft == 0) {
        return strcmp(current, target) == 0 ? 1 : 0;
    }
    
    int cached = findMemo(current, operationsLeft);
    if (cached != -1) {
        return cached;
    }
    
    long long count = 0;
    char* newString = malloc((n + 1) * sizeof(char));
    
    // Try all possible suffix lengths from 1 to n-1
    for (int suffixLen = 1; suffixLen < n; suffixLen++) {
        // Remove suffix of length suffixLen and append to front
        strcpy(newString, current + n - suffixLen);
        strncat(newString, current, n - suffixLen);
        newString[n] = '\0';
        
        count = (count + dp(newString, target, operationsLeft - 1, n)) % MOD;
    }
    
    free(newString);
    addMemo(current, operationsLeft, count);
    return count;
}

int solution(char* s, char* t, int k) {
    memoSize = 0;
    int n = strlen(s);
    return dp(s, t, k, n);
}

int main() {
    char s[1001], t[1001];
    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);
    
    int result = solution(s, t, k);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n^k × n)

At most n^k unique states, each taking O(n) time for string operations

✓ Linear Growth

Space Complexity

O(n^k × n)

Memo table stores up to n^k states, each string takes O(n) space

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Memoized Recursion — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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