Make Three Strings Equal - Practice Coding Problems

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

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Understanding the Visualization

Input

Three strings of different lengths

Find Common Prefix

Identify the longest prefix shared by all strings

Calculate Operations

Sum characters to remove from each string

Key Takeaway

🎯 Key Insight: The equal strings must be the longest common prefix of all three input strings

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

Brute Force — Algorithm Steps

Step 1: Try each possible prefix length
Step 2: Check if all strings share this prefix
Step 3: Calculate operations needed
Step 4: Return minimum operations

Visualization

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Step-by-Step Walkthrough

Start with longest possible prefix

Check if all strings share the longest possible common prefix

Decrease length and retry

If not equal, try shorter prefixes until finding a match

Calculate operations

Sum up 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;
    
    for (int length = minLen; length >= 1; length--) {
        int match = 1;
        for (int i = 0; i < length; i++) {
            if (s1[i] != s2[i] || s2[i] != s3[i]) {
                match = 0;
                break;
            }
        }
        
        if (match) {
            int operations = (len1 - length) + (len2 - length) + (len3 - length);
            return operations;
        }
    }
    
    return -1;
}

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(n²)

For each length up to n, we compare prefixes of length up to n

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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Medium Frequency

~10 min Avg. Time

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Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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