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926. Flip String to Monotone Increasing

Description

A binary string is monotone increasing if it consists of some number of 0's (possibly none), followed by some number of 1's (also possibly none).

You are given a binary string s. You can flip s[i] changing it from 0 to 1 or from 1 to 0.

Return the minimum number of flips to make s monotone increasing.

 

Example 1:

Input: s = "00110"
Output: 1
Explanation: We flip the last digit to get 00111.

Example 2:

Input: s = "010110"
Output: 2
Explanation: We flip to get 011111, or alternatively 000111.

Example 3:

Input: s = "00011000"
Output: 2
Explanation: We flip to get 00000000.

 

Constraints:

  • 1 <= s.length <= 105
  • s[i] is either '0' or '1'.

Solutions

  • class Solution {
    public int minFlipsMonoIncr(String s) {
        int n = s.length();
        int[] left = new int[n + 1];
        int[] right = new int[n + 1];
        int ans = Integer.MAX_VALUE;
        for (int i = 1; i <= n; i++) {
            left[i] = left[i - 1] + (s.charAt(i - 1) == '1' ? 1 : 0);
        }
        for (int i = n - 1; i >= 0; i--) {
            right[i] = right[i + 1] + (s.charAt(i) == '0' ? 1 : 0);
        }
        for (int i = 0; i <= n; i++) {
            ans = Math.min(ans, left[i] + right[i]);
        }
        return ans;
    }
}


  • class Solution {
public:
    int minFlipsMonoIncr(string s) {
        int n = s.size();
        vector<int> left(n + 1, 0), right(n + 1, 0);
        int ans = INT_MAX;
        for (int i = 1; i <= n; ++i) {
            left[i] = left[i - 1] + (s[i - 1] == '1');
        }
        for (int i = n - 1; i >= 0; --i) {
            right[i] = right[i + 1] + (s[i] == '0');
        }
        for (int i = 0; i <= n; i++) {
            ans = min(ans, left[i] + right[i]);
        }
        return ans;
    }
};


  • class Solution:
    def minFlipsMonoIncr(self, s: str) -> int:
        n = len(s)
        left, right = [0] * (n + 1), [0] * (n + 1)
        ans = 0x3F3F3F3F
        for i in range(1, n + 1):
            left[i] = left[i - 1] + (1 if s[i - 1] == '1' else 0)
        for i in range(n - 1, -1, -1):
            right[i] = right[i + 1] + (1 if s[i] == '0' else 0)
        for i in range(0, n + 1):
            ans = min(ans, left[i] + right[i])
        return ans


  • func minFlipsMonoIncr(s string) int {
	n := len(s)
	left, right := make([]int, n+1), make([]int, n+1)
	ans := math.MaxInt32
	for i := 1; i <= n; i++ {
		left[i] = left[i-1]
		if s[i-1] == '1' {
			left[i]++
		}
	}
	for i := n - 1; i >= 0; i-- {
		right[i] = right[i+1]
		if s[i] == '0' {
			right[i]++
		}
	}
	for i := 0; i <= n; i++ {
		ans = min(ans, left[i]+right[i])
	}
	return ans
}

  • /**
 * @param {string} s
 * @return {number}
 */
var minFlipsMonoIncr = function (s) {
    const n = s.length;
    let presum = new Array(n + 1).fill(0);
    for (let i = 0; i < n; ++i) {
        presum[i + 1] = presum[i] + (s[i] == '1');
    }
    let ans = presum[n];
    for (let i = 0; i < n; ++i) {
        ans = Math.min(ans, presum[i] + n - i - (presum[n] - presum[i]));
    }
    return ans;
};


  • function minFlipsMonoIncr(s: string): number {
    let tot = 0;
    for (const c of s) {
        tot += c === '0' ? 1 : 0;
    }
    let [ans, cur] = [tot, 0];
    for (let i = 1; i <= s.length; ++i) {
        cur += s[i - 1] === '0' ? 1 : 0;
        ans = Math.min(ans, i - cur + tot - cur);
    }
    return ans;
}

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