子数组的最小值之和

/*
 * 思路：
 * 1. 由于Java Stream对处理这种复杂的子数组最小值求和不太适用，
 *    这里我们仍然使用单调栈的方法来实现，但用stream来简化代码结构。
 */

import java.util.*;
import java.util.stream.IntStream;

public class Solution {
    public int sumSubarrayMins(int[] arr) {
        int MOD = 1_000_000_007;
        int n = arr.length;
        int[] left = new int[n];
        int[] right = new int[n];
        Stack<Integer> stack = new Stack<>();

        // Fill left array using stream
        IntStream.range(0, n).forEach(i -> {
            while (!stack.isEmpty() && arr[stack.peek()] > arr[i]) {
                stack.pop();
            }
            left[i] = stack.isEmpty() ? -1 : stack.peek();
            stack.push(i);
        });

        // Clear the stack for reuse
        stack.clear();

        // Fill right array using stream
        IntStream.iterate(n - 1, i -> i >= 0, i -> i - 1).forEach(i -> {
            while (!stack.isEmpty() && arr[stack.peek()] >= arr[i]) {
                stack.pop();
            }
            right[i] = stack.isEmpty() ? n : stack.peek();
            stack.push(i);
        });

        // Calculate the result using stream
        return (int) IntStream.range(0, n)
                .mapToLong(i -> (long) arr[i] * (i - left[i]) * (right[i] - i) % MOD)
                .reduce(0, (a, b) -> (a + b) % MOD);
    }
}