第k大二分查找1139.第k大的子数组

    public long thekthSubarray(int[] a, long k) {
        long l = 0;
        long r = 0;
        for (int num : a) r += num;
        while (r - l > 1) {
            long mid = l + (r - l) / 2;
            if (helper(a, mid) >= k) r = mid;
            else l = mid;
        }
        return r;
    }
    
    private long helper(int[] nums, long k) {
        int n = nums.length;
        long res = 0;
        int start = 0;
        long sum = 0;
        for (int end = 0; end  k) {
                res += n - end;
                sum -= nums[start++];
                while (sum > k) sum -= nums[end--];
            }
        }
        return (long)n * (n + 1) / 2 - res;
    }