LeetCode 1737. Change Minimum Characters to Satisfy One of Three Conditions Solution in Java, C++, Python & More | Explanation + Code

Description

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.

Solutions

Solution 1: Counting + Enumeration

First, we count the number of occurrences of each letter in strings a and b, denoted as cnt₁ and cnt₂.

Then, we consider condition 3, i.e., every letter in a and b is the same. We just need to enumerate the final letter c, and then count the number of letters in a and b that are not c. This is the number of characters that need to be changed.

Next, we consider conditions 1 and 2, i.e., every letter in a is less than every letter in b, or every letter in b is less than every letter in a. For condition 1, we make all characters in string a less than character c, and all characters in string b not less than c. We enumerate c to find the smallest answer. Condition 2 is similar.

The final answer is the minimum of the above three cases.

The time complexity is O(m + n + C²), where m and n are the lengths of strings a and b respectively, and C is the size of the character set. In this problem, C = 26.

PythonJavaC++GoTypeScript

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(code-box)

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); } } }(code-box)

class Solution { public: int minCharacters(string a, string b) { int m = a.size(), n = b.size(); vector<int> cnt1(26); vector<int> cnt2(26); for (char& c : a) ++cnt1[c - 'a']; for (char& c : b) ++cnt2[c - 'a']; int ans = m + n; for (int i = 0; i < 26; ++i) ans = min(ans, m + n - cnt1[i] - cnt2[i]); auto f = [&](vector<int>& cnt1, vector<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 = min(ans, t); } }; f(cnt1, cnt2); f(cnt2, cnt1); return ans; } };(code-box)

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 }(code-box)

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; }(code-box)