Make Three Strings Equal

Make Three Strings Equal - Problem

String Easy

You are given three strings: s1, s2, and s3. In one operation you can choose one of these strings and delete its rightmost character.

Note: You cannot completely empty a string.

Return the minimum number of operations required to make the strings equal. If it is impossible to make them equal, return -1.

Input & Output

Example 1 — Basic Case

$ Input: s1 = "abc", s2 = "abcd", s3 = "ab"

› Output: 3

💡 Note: The longest common prefix is "ab". Remove 1 character from s1, 2 from s2, and 0 from s3. Total: 1 + 2 + 0 = 3 operations.

Example 2 — No Common Prefix

$ Input: s1 = "abc", s2 = "def", s3 = "ghi"

› Output: -1

💡 Note: No common prefix exists, so it's impossible to make them equal.

Example 3 — Already Equal

$ Input: s1 = "test", s2 = "test", s3 = "test"

› Output: 0

💡 Note: All strings are already equal, so no operations are needed.

Constraints

1 ≤ s1.length, s2.length, s3.length ≤ 100
s1, s2, and s3 consist only of lowercase English letters

Visualization

Tap to expand

Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is that the final equal strings must be a common prefix of all three input strings. Best approach finds the longest common prefix in O(n) time by comparing characters position by position. Time: O(n), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Optimized Single Pass	O(n)	O(1)	Find the longest common prefix in one pass and calculate operations
Brute Force	O(n²)	O(1)	Try all possible prefix lengths and check if they can be made equal

Optimized Single Pass — Algorithm Steps

Step 1: Compare characters at each position simultaneously
Step 2: Stop at first mismatch or end of shortest string
Step 3: Calculate operations as sum of excess characters

Visualization

Tap to expand

Step-by-Step Walkthrough

Compare first characters

Check if s1[0], s2[0], s3[0] are equal

Continue until mismatch

Keep comparing characters while they match

Calculate operations

Sum characters to remove from each string

Code -

solution.c — C

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

int solution(char* s1, char* s2, char* s3) {
    int len1 = strlen(s1);
    int len2 = strlen(s2);
    int len3 = strlen(s3);
    int minLen = len1;
    if (len2 < minLen) minLen = len2;
    if (len3 < minLen) minLen = len3;
    
    int commonLen = 0;
    for (int i = 0; i < minLen; i++) {
        if (s1[i] == s2[i] && s2[i] == s3[i]) {
            commonLen++;
        } else {
            break;
        }
    }
    
    if (commonLen == 0) {
        return -1;
    }
    
    int operations = (len1 - commonLen) + (len2 - commonLen) + (len3 - commonLen);
    return operations;
}

int main() {
    char s1[1001], s2[1001], s3[1001];
    fgets(s1, sizeof(s1), stdin);
    fgets(s2, sizeof(s2), stdin);
    fgets(s3, sizeof(s3), stdin);
    
    // Remove newlines
    s1[strcspn(s1, "\n")] = 0;
    s2[strcspn(s2, "\n")] = 0;
    s3[strcspn(s3, "\n")] = 0;
    
    int result = solution(s1, s2, s3);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through strings up to the shortest length

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

8.5K Views

Medium Frequency

~10 min Avg. Time

340 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Optimized Single Pass — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler