Replace the Substring for Balanced String

Description

You are given a string s of length n containing only four kinds of characters: 'Q', 'W', 'E', and 'R'.

A string is said to be balanced if each of its characters appears n / 4 times where n is the length of the string.

Return the minimum length of the substring that can be replaced with any other string of the same length to make s balanced. If s is already balanced, return 0.

Example 1:

Input: s = “QWER” Output: 0 Explanation: s is already balanced.

Example 2:

Input: s = “QQWE” Output: 1 Explanation: We need to replace a ‘Q’ to ‘R’, so that “RQWE” (or “QRWE”) is balanced.

Example 3:

Input: s = “QQQW” Output: 2 Explanation: We can replace the first “QQ” to “ER”.

Constraints:

• n == s.length
• 4 <= n <= 105
• n is a multiple of 4.
• s contains only 'Q', 'W', 'E', and 'R'.

Code

Sliding Window

Time Complexity: $O(N)$, Space Complexity: $O(1)$

因為我們可以任意更改 sliding window 中的 element，因此只要 window 外的元素每個的總數都不超過 n / 4，我們就可以更改 window 中的 element 去增加其總數，使得 string 是 balanced 。但是若 window 外的某元素的個數超過 n / 4，我們怎麼改 window 中的 element 都無法使其總數減少，string 就無法被 balanced。

和 Minimum Size Subarray Sum 一樣，在每一輪的 sliding window 結束後，要檢查條件是否滿足（if(count['Q'] <= k && count['W'] <= k && count['E'] <= k && count['R'] <= k) ），而非去檢查邊界條件（j < n），再更新解(res)。

class Solution {
public:
    int balancedString(string s) {
        unordered_map<char, int> count;
        for(auto c: s) {
            count[c]++;
        }
        int n = s.length(), j = 0;
        int k = n / 4;
        int res = n;
        
        for(int i = 0; i < n; i++) {
            while(j < n && (count['Q'] > k || count['W'] > k || count['E'] > k || count['R'] > k)) {
                count[s[j]]--;
                j++;
            }
            if(count['Q'] <= k && count['W'] <= k && count['E'] <= k && count['R'] <= k) {
                res = min(res, j - i);
            }
            count[s[i]]++;
        }
        return res;
    }
 
};

class Solution {
public:
    int balancedString(string s) {
        unordered_map<char, int> count;
        for(auto c: s) {
            count[c]++;
        }
 
        int n = s.length();
        int k = n / 4;
        int i = 0, res = n;
        for(int j = 0; j < n; j++) {
            count[s[j]]--;
            while(i < n && count['Q'] <= k && count['W'] <= k && count['E'] <= k && count['R'] <= k) {
                res = min(res, j - i + 1);
                count[s[i++]]++;
            }
        }
        return res;
 
    }
 
};

Source

• Replace the Substring for Balanced String - LeetCode