Sub-array and question

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Poorly prefix hash,
up to sliding window
---鲁迅

Prefix hash

560. The sub-array whose sum is K (count the number of sub-arrays)

560. Sub-array whose sum is K

Because there is no constraint of positive integers, sliding windows cannot be used

Prefix and + hash table

class Solution:
    def subarraySum(self, nums: List[int], k: int) -> int:
        mp=collections.defaultdict(int)
        mp[0]=1
        pre=0
        count=0
        for num in nums:
            pre+=num
            if pre-k in mp:
                count+=mp[pre-k]
            mp[pre]+=1
        return count

325. The length of the longest sub-array whose sum is equal to k (the maximum length of the statistical sub-array)

325. The length of the longest subarray whose sum is equal to k

class Solution:
    def maxSubArrayLen(self, nums: List[int], k: int) -> int:
        n=len(nums)
        # 上题存个数，这题存下标
        pre_idx = defaultdict(int)
        res=0
        pre=0
        # 忘了写这个
        pre_idx[0] = -1
        for i in range(n):
            pre+=nums[i]
            if pre not in pre_idx:
                pre_idx[pre]=i
            if pre-k in pre_idx:
                res=max(res, i-pre_idx[pre-k])
        return res

523. Contiguous sub-array sum

523. Contiguous sub-array sum

Free to watch

Sliding window

209. The smallest sub-array

209. The smallest sub-array

There are positive integer constraints, using sliding window , $O\left(n\right)$

But you can also use prefix and + dichotomy, $O(n\log n)$

class Solution {
    
    
public:
    int minSubArrayLen(int target, vector<int> &nums) {
    
    
        int sum = 0, l = 0, sz = nums.size() + 1;
        for (int r = 0; r < nums.size(); ++r) {
    
    
            sum += nums[r];
            while (sum >= target) {
    
    
                int cur_sz = r - l + 1;
                if (cur_sz < sz) sz = cur_sz;
                sum -= nums[l++];
            }
        }
        return sz == nums.size() + 1 ? 0 : sz;
    }
};