209. Minimum Size Subarray Sum

这篇文章讨论了如何找到给定正整数数组中,满足和大于或等于目标值的最小子数组长度。通过迭代和优化,Solution类提供了求解该问题的函数。

Given an array of positive integers nums and a positive integer target, return the minimal length of a 

subarray

 whose sum is greater than or equal to target. If there is no such subarray, return 0 instead.

Example 1:

Input: target = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: The subarray [4,3] has the minimal length under the problem constraint.

Example 2:

Input: target = 4, nums = [1,4,4]
Output: 1

Example 3:

Input: target = 11, nums = [1,1,1,1,1,1,1,1]
Output: 0
class Solution {
public:
    int minSubArrayLen(int target, vector<int>& nums) {
        int sum = 0;
        int ans = 0;
        int i = 0;
        int result = INT32_MAX;
        for(int j = 0;j < nums.size();j ++){
            sum+=nums[j];
            while(sum>= target){
                result = min(j - i + 1,result);
                sum -= nums[i];
                i++;
            }
        }
        if(result == INT32_MAX)
            return 0;
        return result;

    }
};

This problem can be solved using binary search. We can try to find the minimum possible maximum sum in a subarray by binary searching over the range of possible values. The upper bound of this range is the sum of all elements in the array, since each subarray must have at least one element. The lower bound of this range is the maximum element in the array, since each subarray must have at least one element and the maximum element must be in its own subarray. For each guess of the maximum sum, we can try to divide the array into subarrays such that no subarray has a sum greater than the guess. This can be done by iterating through the array and greedily assigning each element to the current subarray until the sum of the subarray is greater than the guess. Then, we start a new subarray with the current element. If we can divide the array into k subarrays with a maximum sum no greater than our guess, we can try a smaller guess. If we cannot divide the array into k subarrays with a maximum sum no greater than our guess, we need to try a larger guess. Here's some sample code in Python: ``` n, k = map(int, input().split()) arr = list(map(int, input().split())) low = max(arr) high = sum(arr) while low < high: mid = (low + high) // 2 count = 1 total = 0 for x in arr: if total + x > mid: count += 1 total = x else: total += x if count > k: low = mid + 1 else: high = mid print(low) ``` This code reads in the input and initializes the range of possible values for the maximum sum in a subarray. Then, it performs binary search to find the minimum possible maximum sum. For each guess of the maximum sum, it tries to divide the array into k subarrays such that no subarray has a sum greater than the guess. If it can do this, it tries a smaller guess. If not, it tries a larger guess. Finally, it prints out the minimum possible maximum sum.
评论
成就一亿技术人!
拼手气红包6.0元
还能输入1000个字符
 
红包 添加红包
表情包 插入表情
 条评论被折叠 查看
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值