使数组严格递增

/*
 * Problem: Given two integer arrays arr1 and arr2, return the minimum number of operations required to make arr1 strictly increasing by replacing elements from arr1 with elements from arr2. If it is impossible, return -1.
 * Approach: Use dynamic programming and Java Streams to process the arrays and keep track of the minimum operations required for each possible ending value of arr1 up to each index.
 */
import java.util.Arrays;
import java.util.stream.IntStream;

public class Solution {
    public int makeArrayIncreasing(int[] arr1, int[] arr2) {
        Arrays.sort(arr2);
        int n = arr1.length;
        int m = arr2.length;
        int[][] dp = new int[n + 1][m + 1];
        for (int[] row : dp) Arrays.fill(row, Integer.MAX_VALUE);
        dp[0][0] = 0;

        IntStream.range(1, n + 1).forEach(i -> {
            IntStream.rangeClosed(0, m).forEach(j -> {
                if (j == 0 || arr2[j - 1] < arr1[i - 1]) {
                    dp[i][0] = Math.min(dp[i][0], dp[i - 1][j]);
                }
                if (j > 0 && arr2[j - 1] < arr1[i - 1]) {
                    dp[i][j] = Math.min(dp[i][j], dp[i - 1][0] + 1);
                }
                if (j > 0 && arr2[j - 1] < arr2[j - 2]) {
                    dp[i][j] = Math.min(dp[i][j], dp[i - 1][j - 1] + 1);
                }
            });
        });

        return IntStream.rangeClosed(0, m).map(j -> dp[n][j]).min().orElse(Integer.MAX_VALUE) == Integer.MAX_VALUE ? -1 : IntStream.rangeClosed(0, m).map(j -> dp[n][j]).min().orElse(Integer.MAX_VALUE);
    }
}