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package class_2022_01_3_week;
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// 来自字节跳动
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// 给定一个数组arr，其中的值有可能正、负、0
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// 给定一个正数k
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// 返回累加和>=k的所有子数组中，最短的子数组长度
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// 本题测试链接 : https://leetcode.com/problems/shortest-subarray-with-sum-at-least-k/
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public class Code03_ShortestSubarrayWithSumAtLeastK {
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public static int shortestSubarray1(int[] arr, int k) {
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if (arr == null || arr.length < 1) {
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return -1;
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}
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int n = arr.length + 1;
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long[] sum = new long[n];
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for (int i = 1; i < n; i++) {
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sum[i] = sum[i - 1] + arr[i - 1];
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}
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int[] stack = new int[n];
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int size = 1;
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int ans = Integer.MAX_VALUE;
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for (int i = 1; i < n; i++) {
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int mostRight = mostRight(sum, stack, size, sum[i] - k);
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ans = Math.min(ans, mostRight == -1 ? Integer.MAX_VALUE : (i - mostRight));
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while (size > 0 && sum[stack[size - 1]] >= sum[i]) {
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size--;
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}
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stack[size++] = i;
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}
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return ans == Integer.MAX_VALUE ? -1 : ans;
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}
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public static int mostRight(long[] sum, int[] stack, int size, long aim) {
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int l = 0;
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int r = size - 1;
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int m = 0;
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int ans = -1;
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while (l <= r) {
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m = (l + r) / 2;
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if (sum[stack[m]] <= aim) {
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ans = stack[m];
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l = m + 1;
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} else {
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r = m - 1;
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}
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}
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return ans;
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}
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public static int shortestSubarray2(int[] arr, int K) {
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int N = arr.length;
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long[] sum = new long[N + 1];
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for (int i = 0; i < N; i++) {
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sum[i + 1] = sum[i] + arr[i];
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}
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int ans = Integer.MAX_VALUE;
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int[] dq = new int[N + 1];
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int l = 0;
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int r = 0;
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for (int i = 0; i < N + 1; i++) {
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// 头部开始，符合条件的，从头部弹出！
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while (l != r && sum[i] - sum[dq[l]] >= K) {
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ans = Math.min(ans, i - dq[l++]);
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}
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// 尾部开始，前缀和比当前的前缀和大于等于的，从尾部弹出！
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while (l != r && sum[dq[r - 1]] >= sum[i]) {
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r--;
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}
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dq[r++] = i;
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}
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return ans != Integer.MAX_VALUE ? ans : -1;
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}
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}
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