LeetCode #2896 — MEDIUM

Apply Operations to Make Two Strings Equal

Move from brute-force thinking to an efficient approach using dynamic programming strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given two 0-indexed binary strings s1 and s2, both of length n, and a positive integer x.

You can perform any of the following operations on the string s1 any number of times:

Choose two indices i and j, and flip both s1[i] and s1[j]. The cost of this operation is x.
Choose an index i such that i < n - 1 and flip both s1[i] and s1[i + 1]. The cost of this operation is 1.

Return the minimum cost needed to make the strings s1 and s2 equal, or return -1 if it is impossible.

Note that flipping a character means changing it from 0 to 1 or vice-versa.

Example 1:

Input: s1 = "1100011000", s2 = "0101001010", x = 2
Output: 4
Explanation: We can do the following operations:
- Choose i = 3 and apply the second operation. The resulting string is s1 = "1101111000".
- Choose i = 4 and apply the second operation. The resulting string is s1 = "1101001000".
- Choose i = 0 and j = 8 and apply the first operation. The resulting string is s1 = "0101001010" = s2.
The total cost is 1 + 1 + 2 = 4. It can be shown that it is the minimum cost possible.

Example 2:

Input: s1 = "10110", s2 = "00011", x = 4
Output: -1
Explanation: It is not possible to make the two strings equal.

Constraints:

n == s1.length == s2.length
1 <= n, x <= 500
s1 and s2 consist only of the characters '0' and '1'.

Step 01

Brute Force Baseline

Problem summary: You are given two 0-indexed binary strings s1 and s2, both of length n, and a positive integer x. You can perform any of the following operations on the string s1 any number of times: Choose two indices i and j, and flip both s1[i] and s1[j]. The cost of this operation is x. Choose an index i such that i < n - 1 and flip both s1[i] and s1[i + 1]. The cost of this operation is 1. Return the minimum cost needed to make the strings s1 and s2 equal, or return -1 if it is impossible. Note that flipping a character means changing it from 0 to 1 or vice-versa.

Baseline thinking

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

Pattern signal: Dynamic Programming

Example 1

"1100011000"
"0101001010"
2

Example 2

"10110"
"00011"
4

Step 02

Core Insight

What unlocks the optimal approach

Save all the indices that have different characters on <code>s1</code> and <code>s2</code> into a list, and work only with this list.
Try to use dynamic programming on this list to solve the problem. What will be the states and transitions of this dp?

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 #2896: Apply Operations to Make Two Strings Equal
class Solution {
    private List<Integer> idx = new ArrayList<>();
    private Integer[][] f;
    private int x;

    public int minOperations(String s1, String s2, int x) {
        int n = s1.length();
        for (int i = 0; i < n; ++i) {
            if (s1.charAt(i) != s2.charAt(i)) {
                idx.add(i);
            }
        }
        int m = idx.size();
        if (m % 2 == 1) {
            return -1;
        }
        this.x = x;
        f = new Integer[m][m];
        return dfs(0, m - 1);
    }

    private int dfs(int i, int j) {
        if (i > j) {
            return 0;
        }
        if (f[i][j] != null) {
            return f[i][j];
        }
        f[i][j] = dfs(i + 1, j - 1) + x;
        f[i][j] = Math.min(f[i][j], dfs(i + 2, j) + idx.get(i + 1) - idx.get(i));
        f[i][j] = Math.min(f[i][j], dfs(i, j - 2) + idx.get(j) - idx.get(j - 1));
        return f[i][j];
    }
}

// Accepted solution for LeetCode #2896: Apply Operations to Make Two Strings Equal
func minOperations(s1 string, s2 string, x int) int {
	idx := []int{}
	for i := range s1 {
		if s1[i] != s2[i] {
			idx = append(idx, i)
		}
	}
	m := len(idx)
	if m&1 == 1 {
		return -1
	}
	f := make([][]int, m)
	for i := range f {
		f[i] = make([]int, m)
		for j := range f[i] {
			f[i][j] = -1
		}
	}
	var dfs func(i, j int) int
	dfs = func(i, j int) int {
		if i > j {
			return 0
		}
		if f[i][j] != -1 {
			return f[i][j]
		}
		f[i][j] = dfs(i+1, j-1) + x
		f[i][j] = min(f[i][j], dfs(i+2, j)+idx[i+1]-idx[i])
		f[i][j] = min(f[i][j], dfs(i, j-2)+idx[j]-idx[j-1])
		return f[i][j]
	}
	return dfs(0, m-1)
}

# Accepted solution for LeetCode #2896: Apply Operations to Make Two Strings Equal
class Solution:
    def minOperations(self, s1: str, s2: str, x: int) -> int:
        @cache
        def dfs(i: int, j: int) -> int:
            if i > j:
                return 0
            a = dfs(i + 1, j - 1) + x
            b = dfs(i + 2, j) + idx[i + 1] - idx[i]
            c = dfs(i, j - 2) + idx[j] - idx[j - 1]
            return min(a, b, c)

        n = len(s1)
        idx = [i for i in range(n) if s1[i] != s2[i]]
        m = len(idx)
        if m & 1:
            return -1
        return dfs(0, m - 1)

// Accepted solution for LeetCode #2896: Apply Operations to Make Two Strings Equal
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2896: Apply Operations to Make Two Strings Equal
// class Solution {
//     private List<Integer> idx = new ArrayList<>();
//     private Integer[][] f;
//     private int x;
// 
//     public int minOperations(String s1, String s2, int x) {
//         int n = s1.length();
//         for (int i = 0; i < n; ++i) {
//             if (s1.charAt(i) != s2.charAt(i)) {
//                 idx.add(i);
//             }
//         }
//         int m = idx.size();
//         if (m % 2 == 1) {
//             return -1;
//         }
//         this.x = x;
//         f = new Integer[m][m];
//         return dfs(0, m - 1);
//     }
// 
//     private int dfs(int i, int j) {
//         if (i > j) {
//             return 0;
//         }
//         if (f[i][j] != null) {
//             return f[i][j];
//         }
//         f[i][j] = dfs(i + 1, j - 1) + x;
//         f[i][j] = Math.min(f[i][j], dfs(i + 2, j) + idx.get(i + 1) - idx.get(i));
//         f[i][j] = Math.min(f[i][j], dfs(i, j - 2) + idx.get(j) - idx.get(j - 1));
//         return f[i][j];
//     }
// }

// Accepted solution for LeetCode #2896: Apply Operations to Make Two Strings Equal
function minOperations(s1: string, s2: string, x: number): number {
    const idx: number[] = [];
    for (let i = 0; i < s1.length; ++i) {
        if (s1[i] !== s2[i]) {
            idx.push(i);
        }
    }
    const m = idx.length;
    if (m % 2 === 1) {
        return -1;
    }
    if (m === 0) {
        return 0;
    }
    const f: number[][] = Array.from({ length: m }, () => Array.from({ length: m }, () => -1));
    const dfs = (i: number, j: number): number => {
        if (i > j) {
            return 0;
        }
        if (f[i][j] !== -1) {
            return f[i][j];
        }
        f[i][j] = dfs(i + 1, j - 1) + x;
        f[i][j] = Math.min(f[i][j], dfs(i + 2, j) + idx[i + 1] - idx[i]);
        f[i][j] = Math.min(f[i][j], dfs(i, j - 2) + idx[j] - idx[j - 1]);
        return f[i][j];
    };
    return dfs(0, m - 1);
}

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^2)

Space

O(n^2)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.