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Max Size Subarray Sum Equals k

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Description

https://leetcode.com/problems/maximum-size-subarray-sum-equals-k/

Brutal Force O(n^3)

Prefix Sum: O(n^2)

Naive Hash Table: O(n)

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class Solution {
public:
    int maxSubArrayLen(vector<int>& nums, int k) {
        int ret = 0;
        int sum = 0;
        int size = nums.size();
        if (size == 0) return 0;
        unordered_map<int, vector<int>> mapping; ///Mapping the sum to [0,..index]
        vector<int> dp(size + 1, -1);
        for (int i = 0; i < size; ++i) {
            dp[i] = sum;
            sum += nums[i];
            mapping[dp[i]].push_back(i);
        }
        dp[size] = sum;
        sum += nums[size];
        mapping[dp[size]].push_back(size); 
        
        for (int j = 0; j < size + 1; ++j) {
            if (mapping.count(dp[j] - k)) {
                for (auto& item : mapping[dp[j] - k])
                    ret = max(ret, j - item);
            }
        }
        
        return ret;
    }
};

Hash Table: O(n)

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class Solution {
public:
    int maxSubArrayLen(vector<int>& nums, int k) {
        int ret = 0;
        int sum = 0;
        int size = nums.size();
        if (size == 0) return 0;
        unordered_map<int, int> mapping; ///Mapping the sum to [0,..index]
        vector<int> dp(size + 1, -1);
        dp[0] = 0;
        mapping[0] = 0;
        for (int i = 1; i <= size; ++i) {
            sum += nums[i - 1];
            dp[i] = sum;
            if (mapping.count(dp[i]) == 0) mapping[dp[i]] = i;
        }
        
        for (int j = 0; j < size + 1; ++j) {
            if (mapping.count(dp[j] - k)) {
                ret = max(ret, j - mapping[dp[j] - k]);
            }
        }
        
        return ret;
    }
};