LeetCode #1737 — MEDIUM

Change Minimum Characters to Satisfy One of Three Conditions

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

Problem Statement

You are given two strings a and b that consist of lowercase letters. In one operation, you can change any character in a or b to any lowercase letter.

Your goal is to satisfy one of the following three conditions:

Every letter in a is strictly less than every letter in b in the alphabet.
Every letter in b is strictly less than every letter in a in the alphabet.
Both a and b consist of only one distinct letter.

Return the minimum number of operations needed to achieve your goal.

Example 1:

Input: a = "aba", b = "caa"
Output: 2
Explanation: Consider the best way to make each condition true:
1) Change b to "ccc" in 2 operations, then every letter in a is less than every letter in b.
2) Change a to "bbb" and b to "aaa" in 3 operations, then every letter in b is less than every letter in a.
3) Change a to "aaa" and b to "aaa" in 2 operations, then a and b consist of one distinct letter.
The best way was done in 2 operations (either condition 1 or condition 3).

Example 2:

Input: a = "dabadd", b = "cda"
Output: 3
Explanation: The best way is to make condition 1 true by changing b to "eee".

Constraints:

1 <= a.length, b.length <= 10⁵
a and b consist only of lowercase letters.

Step 01

Brute Force Baseline

Problem summary: You are given two strings a and b that consist of lowercase letters. In one operation, you can change any character in a or b to any lowercase letter. Your goal is to satisfy one of the following three conditions: Every letter in a is strictly less than every letter in b in the alphabet. Every letter in b is strictly less than every letter in a in the alphabet. Both a and b consist of only one distinct letter. Return the minimum number of operations needed to achieve your goal.

Baseline thinking

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

Pattern signal: Hash Map

Example 1

"aba"
"caa"

Example 2

"dabadd"
"cda"

Step 02

Core Insight

What unlocks the optimal approach

Iterate on each letter in the alphabet, and check the smallest number of operations needed to make it one of the following: the largest letter in a and smaller than the smallest one in b, vice versa, or let a and b consist only of this letter.
For the first 2 conditions, take care that you can only change characters to lowercase letters, so you can't make 'z' the smallest letter in one of the strings or 'a' the largest letter in one of them.

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 #1737: Change Minimum Characters to Satisfy One of Three Conditions
class Solution {
    private int ans;

    public int minCharacters(String a, String b) {
        int m = a.length(), n = b.length();
        int[] cnt1 = new int[26];
        int[] cnt2 = new int[26];
        for (int i = 0; i < m; ++i) {
            ++cnt1[a.charAt(i) - 'a'];
        }
        for (int i = 0; i < n; ++i) {
            ++cnt2[b.charAt(i) - 'a'];
        }
        ans = m + n;
        for (int i = 0; i < 26; ++i) {
            ans = Math.min(ans, m + n - cnt1[i] - cnt2[i]);
        }
        f(cnt1, cnt2);
        f(cnt2, cnt1);
        return ans;
    }

    private void f(int[] cnt1, int[] cnt2) {
        for (int i = 1; i < 26; ++i) {
            int t = 0;
            for (int j = i; j < 26; ++j) {
                t += cnt1[j];
            }
            for (int j = 0; j < i; ++j) {
                t += cnt2[j];
            }
            ans = Math.min(ans, t);
        }
    }
}

// Accepted solution for LeetCode #1737: Change Minimum Characters to Satisfy One of Three Conditions
func minCharacters(a string, b string) int {
	cnt1 := [26]int{}
	cnt2 := [26]int{}
	for _, c := range a {
		cnt1[c-'a']++
	}
	for _, c := range b {
		cnt2[c-'a']++
	}
	m, n := len(a), len(b)
	ans := m + n
	for i := 0; i < 26; i++ {
		ans = min(ans, m+n-cnt1[i]-cnt2[i])
	}
	f := func(cnt1, cnt2 [26]int) {
		for i := 1; i < 26; i++ {
			t := 0
			for j := i; j < 26; j++ {
				t += cnt1[j]
			}
			for j := 0; j < i; j++ {
				t += cnt2[j]
			}
			ans = min(ans, t)
		}
	}
	f(cnt1, cnt2)
	f(cnt2, cnt1)
	return ans
}

# Accepted solution for LeetCode #1737: Change Minimum Characters to Satisfy One of Three Conditions
class Solution:
    def minCharacters(self, a: str, b: str) -> int:
        def f(cnt1, cnt2):
            for i in range(1, 26):
                t = sum(cnt1[i:]) + sum(cnt2[:i])
                nonlocal ans
                ans = min(ans, t)

        m, n = len(a), len(b)
        cnt1 = [0] * 26
        cnt2 = [0] * 26
        for c in a:
            cnt1[ord(c) - ord('a')] += 1
        for c in b:
            cnt2[ord(c) - ord('a')] += 1
        ans = m + n
        for c1, c2 in zip(cnt1, cnt2):
            ans = min(ans, m + n - c1 - c2)
        f(cnt1, cnt2)
        f(cnt2, cnt1)
        return ans

// Accepted solution for LeetCode #1737: Change Minimum Characters to Satisfy One of Three Conditions
struct Solution;

impl Solution {
    fn min_characters(a: String, b: String) -> i32 {
        let a: Vec<u8> = a.bytes().collect();
        let b: Vec<u8> = b.bytes().collect();
        let count_a = Self::check(&a);
        let count_b = Self::check(&b);
        let mut prefix_a = vec![0; 26];
        let mut prev_a = 0;
        let mut prefix_b = vec![0; 26];
        let mut prev_b = 0;
        for i in 0..26 {
            prefix_a[i] = prev_a;
            prev_a += count_a[i];
            prefix_b[i] = prev_b;
            prev_b += count_b[i];
        }
        let mut postfix_a = vec![0; 26];
        let mut next_a = 0;
        let mut postfix_b = vec![0; 26];
        let mut next_b = 0;
        for i in (0..26).rev() {
            postfix_a[i] = next_a;
            next_a += count_a[i];
            postfix_b[i] = next_b;
            next_b += count_b[i];
        }
        let mut max = 0;
        let mut max_a = 0;
        let mut max_b = 0;
        for i in 0..26 {
            max_a = max_a.max(count_a[i]);
            max_b = max_b.max(count_b[i]);
        }
        max = max.max(max_a + max_b);
        for i in 1..26 {
            max = max.max(prefix_a[i] + postfix_b[i - 1]);
            max = max.max(prefix_b[i] + postfix_a[i - 1]);
        }
        (a.len() + b.len() - max) as i32
    }

    fn check(s: &[u8]) -> Vec<usize> {
        let mut count: Vec<usize> = vec![0; 26];
        for b in s {
            count[(b - b'a') as usize] += 1;
        }
        count
    }
}

#[test]
fn test() {
    let a = "aba".to_string();
    let b = "caa".to_string();
    let res = 2;
    assert_eq!(Solution::min_characters(a, b), res);
    let a = "dabadd".to_string();
    let b = "cda".to_string();
    let res = 3;
    assert_eq!(Solution::min_characters(a, b), res);
    let a = "acac".to_string();
    let b = "bd".to_string();
    let res = 1;
    assert_eq!(Solution::min_characters(a, b), res);
}

// Accepted solution for LeetCode #1737: Change Minimum Characters to Satisfy One of Three Conditions
function minCharacters(a: string, b: string): number {
    const m = a.length,
        n = b.length;
    let count1 = new Array(26).fill(0);
    let count2 = new Array(26).fill(0);
    const base = 'a'.charCodeAt(0);

    for (let char of a) {
        count1[char.charCodeAt(0) - base]++;
    }
    for (let char of b) {
        count2[char.charCodeAt(0) - base]++;
    }

    let pre1 = 0,
        pre2 = 0;
    let ans = m + n;
    for (let i = 0; i < 25; i++) {
        pre1 += count1[i];
        pre2 += count2[i];
        // case1， case2， case3
        ans = Math.min(ans, m - pre1 + pre2, pre1 + n - pre2, m + n - count1[i] - count2[i]);
    }
    ans = Math.min(ans, m + n - count1[25] - count2[25]);

    return ans;
}

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 the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.