[Daily Coding Problem 331] Minimum Flips to make XY string

You are given a string consisting of the letters x and y, such as xyxxxyxyy. In addition, you have an operation called flip, which changes a single x to y or vice versa.

Determine how many times you would need to apply this operation to ensure that all x's come before all y's. In the preceding example, it suffices to flip the second and sixth characters, so you should return 2.

Solution 1. O(N) dynamic programming

dp[i][j]: the minimum flips needed to properly order the substring up to index i, where the final element ends up taking value j.

public class MinimumFlipToMakeXYString {

    public static int minFlips(String s) {
        int n = s.length();
        int[][] dp = new int[n][2];
        dp[0][0] = (s.charAt(0) == 'x' ? 0 : 1);
        dp[0][1] = (s.charAt(0) == 'y' ? 0 : 1);

        for(int i = 1; i < n; i++) {
            dp[i][0] = dp[i - 1][0] + (s.charAt(i) == 'x' ? 0 : 1);
            dp[i][1] = Math.min(dp[i - 1][0], dp[i - 1][1]) + (s.charAt(i) == 'x' ? 1 : 0);
        }
        return Math.min(dp[n - 1][0], dp[n - 1][1]);
    }

    public static void main(String[] args) {
        String s1 = "yyx";
        String s2 = "xyxxxyxyyy";
        String s3 = "xyxxxyxyyx";
        String s4 = "xyxxxyxyyxxx";
        String s5 = "yyxxx";

        System.out.println(minFlips(s5));
    }
}

Solution 2. O(N) Prefix Sum

An alternative solution is as follows. First, we make a pass over our input to determine the number of y's to the left of each element. We then make a second pass to find the number of x's to the right of each element.

For any solution, there must be some element in our string for which everything to its left gets set to x, and everything to its right gets set to y. As a result, we can simply find the pairwise sum of these lists, and use the element which requires the smallest total number of flips.

public class MinimumFlipToMakeXYString {
    public static int minFlips(String s) {
        int n = s.length();
        int[] leftY = new int[n], rightX = new int[n];
        leftY[0] = 0;
        rightX[n - 1] = 0;

        for(int i = 1; i < n; i++) {
            leftY[i] = leftY[i - 1] + (s.charAt(i - 1) == 'y' ? 1 : 0);
        }
        for(int i = n - 2; i >= 0; i--) {
            rightX[i] = rightX[i + 1] + (s.charAt(i + 1) == 'x' ? 1 : 0);
        }
        int res = n;
        for(int i = 0; i < n; i++) {
            res = Math.min(res, leftY[i] + rightX[i]);
        }
        return res;
    }
}

来源：https://www.cnblogs.com/lz87/p/11735918.html

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