LeetCode #2937 — EASY

Make Three Strings Equal

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

The Problem

Problem Statement

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 that 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.

Example 1:

Input: s1 = "abc", s2 = "abb", s3 = "ab"

Output: 2

Explanation: Deleting the rightmost character from both s1 and s2 will result in three equal strings.

Example 2:

Input: s1 = "dac", s2 = "bac", s3 = "cac"

Output: -1

Explanation: Since the first letters of s1 and s2 differ, they cannot be made equal.

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: 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 that 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

"abc"
"abb"
"ab"

Example 2

"dac"
"bac"
"cac"

Core Insight

What unlocks the optimal approach

Calculate the length of the longest common prefix of the <code>3</code> strings.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2937: Make Three Strings Equal
class Solution {
    public int findMinimumOperations(String s1, String s2, String s3) {
        int s = s1.length() + s2.length() + s3.length();
        int n = Math.min(Math.min(s1.length(), s2.length()), s3.length());
        for (int i = 0; i < n; ++i) {
            if (!(s1.charAt(i) == s2.charAt(i) && s2.charAt(i) == s3.charAt(i))) {
                return i == 0 ? -1 : s - 3 * i;
            }
        }
        return s - 3 * n;
    }
}

// Accepted solution for LeetCode #2937: Make Three Strings Equal
func findMinimumOperations(s1 string, s2 string, s3 string) int {
	s := len(s1) + len(s2) + len(s3)
	n := min(len(s1), len(s2), len(s3))
	for i := range s1[:n] {
		if !(s1[i] == s2[i] && s2[i] == s3[i]) {
			if i == 0 {
				return -1
			}
			return s - 3*i
		}
	}
	return s - 3*n
}

# Accepted solution for LeetCode #2937: Make Three Strings Equal
class Solution:
    def findMinimumOperations(self, s1: str, s2: str, s3: str) -> int:
        s = len(s1) + len(s2) + len(s3)
        n = min(len(s1), len(s2), len(s3))
        for i in range(n):
            if not s1[i] == s2[i] == s3[i]:
                return -1 if i == 0 else s - 3 * i
        return s - 3 * n

// Accepted solution for LeetCode #2937: Make Three Strings Equal
fn find_minimum_operations(s1: String, s2: String, s3: String) -> i32 {
    let mut s1 = s1;
    let mut s2 = s2;
    let mut s3 = s3;

    let mut ret = 0;
    let min_len = std::cmp::min(s1.len(), std::cmp::min(s2.len(), s3.len()));

    while s1.len() > min_len {
        s1.pop();
        ret += 1;
    }
    while s2.len() > min_len {
        s2.pop();
        ret += 1;
    }
    while s3.len() > min_len {
        s3.pop();
        ret += 1;
    }

    loop {
        if s1 == s2 && s2 == s3 {
            return ret;
        }

        ret += 3;
        s1.pop();
        s2.pop();
        s3.pop();

        if s1.is_empty() {
            return -1;
        }
    }
}

fn main() {
    let s1 = "abc".to_string();
    let s2 = "abb".to_string();
    let s3 = "ab".to_string();
    let ret = find_minimum_operations(s1, s2, s3);
    println!("ret={ret}");
}

#[test]
fn test() {
    {
        let s1 = "abc".to_string();
        let s2 = "abb".to_string();
        let s3 = "ab".to_string();
        let ret = find_minimum_operations(s1, s2, s3);
        assert_eq!(ret, 2);
    }
    {
        let s1 = "dac".to_string();
        let s2 = "bac".to_string();
        let s3 = "cac".to_string();
        let ret = find_minimum_operations(s1, s2, s3);
        assert_eq!(ret, -1);
    }
    {
        let s1 = "a".to_string();
        let s2 = "a".to_string();
        let s3 = "a".to_string();
        let ret = find_minimum_operations(s1, s2, s3);
        assert_eq!(ret, 0);
    }
    {
        let s1 = "ab".to_string();
        let s2 = "ac".to_string();
        let s3 = "ad".to_string();
        let ret = find_minimum_operations(s1, s2, s3);
        assert_eq!(ret, 3);
    }
    {
        let s1 = "abc".to_string();
        let s2 = "abc".to_string();
        let s3 = "abc".to_string();
        let ret = find_minimum_operations(s1, s2, s3);
        assert_eq!(ret, 0);
    }
}

// Accepted solution for LeetCode #2937: Make Three Strings Equal
function findMinimumOperations(s1: string, s2: string, s3: string): number {
    const s = s1.length + s2.length + s3.length;
    const n = Math.min(s1.length, s2.length, s3.length);
    for (let i = 0; i < n; ++i) {
        if (!(s1[i] === s2[i] && s2[i] === s3[i])) {
            return i === 0 ? -1 : s - 3 * i;
        }
    }
    return s - 3 * n;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.