LeetCode #1544 — EASY

Make The String Great

Build confidence with an intuition-first walkthrough focused on stack fundamentals.

The Problem

Problem Statement

Given a string s of lower and upper case English letters.

A good string is a string which doesn't have two adjacent characters s[i] and s[i + 1] where:

0 <= i <= s.length - 2
s[i] is a lower-case letter and s[i + 1] is the same letter but in upper-case or vice-versa.

To make the string good, you can choose two adjacent characters that make the string bad and remove them. You can keep doing this until the string becomes good.

Return the string after making it good. The answer is guaranteed to be unique under the given constraints.

Notice that an empty string is also good.

Example 1:

Input: s = "leEeetcode"
Output: "leetcode"
Explanation: In the first step, either you choose i = 1 or i = 2, both will result "leEeetcode" to be reduced to "leetcode".

Example 2:

Input: s = "abBAcC"
Output: ""
Explanation: We have many possible scenarios, and all lead to the same answer. For example:
"abBAcC" --> "aAcC" --> "cC" --> ""
"abBAcC" --> "abBA" --> "aA" --> ""

Example 3:

Input: s = "s"
Output: "s"

Constraints:

1 <= s.length <= 100
s contains only lower and upper case English letters.

Step 01

Brute Force Baseline

Problem summary: Given a string s of lower and upper case English letters. A good string is a string which doesn't have two adjacent characters s[i] and s[i + 1] where: 0 <= i <= s.length - 2 s[i] is a lower-case letter and s[i + 1] is the same letter but in upper-case or vice-versa. To make the string good, you can choose two adjacent characters that make the string bad and remove them. You can keep doing this until the string becomes good. Return the string after making it good. The answer is guaranteed to be unique under the given constraints. Notice that an empty string is also good.

Baseline thinking

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

Pattern signal: Stack

Example 1

"leEeetcode"

Example 2

"abBAcC"

Example 3

"s"

Step 02

Core Insight

What unlocks the optimal approach

The order you choose 2 characters to remove doesn't matter.
Keep applying the mentioned step to s till the length of the string is not changed.

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 #1544: Make The String Great
class Solution {
    public String makeGood(String s) {
        StringBuilder sb = new StringBuilder();
        for (char c : s.toCharArray()) {
            if (sb.length() == 0 || Math.abs(sb.charAt(sb.length() - 1) - c) != 32) {
                sb.append(c);
            } else {
                sb.deleteCharAt(sb.length() - 1);
            }
        }
        return sb.toString();
    }
}

// Accepted solution for LeetCode #1544: Make The String Great
func makeGood(s string) string {
	stk := []rune{}
	for _, c := range s {
		if len(stk) == 0 || abs(int(stk[len(stk)-1]-c)) != 32 {
			stk = append(stk, c)
		} else {
			stk = stk[:len(stk)-1]
		}
	}
	return string(stk)
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #1544: Make The String Great
class Solution:
    def makeGood(self, s: str) -> str:
        stk = []
        for c in s:
            if not stk or abs(ord(stk[-1]) - ord(c)) != 32:
                stk.append(c)
            else:
                stk.pop()
        return "".join(stk)

// Accepted solution for LeetCode #1544: Make The String Great
struct Solution;

impl Solution {
    fn make_good(s: String) -> String {
        let mut stack: Vec<char> = vec![];
        for c in s.chars() {
            if let Some(&last) = stack.last() {
                if c.is_uppercase() && last.is_lowercase() && c.to_ascii_lowercase() == last {
                    stack.pop();
                    continue;
                }
                if c.is_lowercase() && last.is_uppercase() && c.to_ascii_uppercase() == last {
                    stack.pop();
                    continue;
                }
            }
            stack.push(c);
        }
        stack.into_iter().collect()
    }
}

#[test]
fn test() {
    let s = "leEeetcode".to_string();
    let res = "leetcode".to_string();
    assert_eq!(Solution::make_good(s), res);
    let s = "abBAcC".to_string();
    let res = "".to_string();
    assert_eq!(Solution::make_good(s), res);
    let s = "s".to_string();
    let res = "s".to_string();
    assert_eq!(Solution::make_good(s), res);
}

// Accepted solution for LeetCode #1544: Make The String Great
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1544: Make The String Great
// class Solution {
//     public String makeGood(String s) {
//         StringBuilder sb = new StringBuilder();
//         for (char c : s.toCharArray()) {
//             if (sb.length() == 0 || Math.abs(sb.charAt(sb.length() - 1) - c) != 32) {
//                 sb.append(c);
//             } else {
//                 sb.deleteCharAt(sb.length() - 1);
//             }
//         }
//         return sb.toString();
//     }
// }

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(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.