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Formatted question description: https://leetcode.ca/all/560.html

560. Subarray Sum Equals K

Level

Medium

Description

Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k.

Example 1:

Input: nums = [1,1,1], k = 2

Output: 2

Note:

  1. The length of the array is in range [1, 20,000].
  2. The range of numbers in the array is [-1000, 1000] and the range of the integer k is [-1e7, 1e7].

Solution

Use a map to store each sum and the count of subarrays of the sum. Put sum 0 with count 1 to the map initially.

Loop over nums and calculate the total sum sum. For each num in nums, add num to sum, and obtain the count of subarrays with sum sum - k. Add the sum to the result. Then update the map with the count of subarrays with sum sum. Finally, return the result.

  • 
public class Subarray_Sum_Equals_K {

    public static void main(String[] args) {
        Subarray_Sum_Equals_K out = new Subarray_Sum_Equals_K();
        Solution s = out.new Solution();

        System.out.println(s.subarraySum(new int[]{1, 1, 1}, 2));
    }

    // time: O(N)
    // space: O(N)

    // https://leetcode.com/problems/subarray-sum-equals-k/solution/
    class Solution_hashmap {
        public int subarraySum(int[] nums, int k) {

            // sum => count of to-this-sum
            HashMap <Integer, Integer> hm = new HashMap<>();
            hm.put(0, 1);

            int count = 0;
            int sum = 0;
            for (int i = 0; i < nums.length; i++) {
                sum += nums[i];
                if (hm.containsKey(sum - k)) { // same idea as, if (sum[end] - sum[start] == k) {
                    count += hm.get(sum - k);
                }

                hm.put(sum, hm.getOrDefault(sum, 0) + 1);
            }

            return count;
        }
    }

    // time: O(N^2)
    // space: O(1)
    class Solution_withNoExtraSpace {
        public int subarraySum(int[] nums, int k) {
            int count = 0;
            for (int start = 0; start < nums.length; start++) {
                int sum = 0; // @note: sum for each start point
                for (int end = start; end < nums.length; end++) {
                    sum += nums[end];
                    if (sum == k)
                        count++;
                }
            }
            return count;
        }
    }

    // time: O(N^2)
    // space: O(N)
    class Solution2 {
        public int subarraySum(int[] nums, int k) {
            int count = 0;

            // sum value from 1 to i-th element
            int[] sum = new int[nums.length + 1];
            sum[0] = 0;

            // cached sum
            for (int i = 1; i <= nums.length; i++) {
                sum[i] = sum[i - 1] + nums[i - 1];
            }

            for (int start = 0; start < nums.length; start++) {
                for (int end = start + 1; end <= nums.length; end++) {
                    if (sum[end] - sum[start] == k) {
                        count++;
                    }
                }
            }
            return count;
        }
    }

    // brute force
    class Solution {
        int count = 0;

        public int subarraySum(int[] nums, int k) {

            for (int start = 0; start < nums.length; start++) {
                for (int end = start + 1; end <= nums.length; end++) {
                    int sum = 0;
                    for (int i = start; i < end; i++) {
                        sum += nums[i];
                    }
                    if (sum == k)
                        count++;
                }
            }

            return count;
        }
    }
}

############

class Solution {
    public int subarraySum(int[] nums, int k) {
        Map<Integer, Integer> counter = new HashMap<>();
        counter.put(0, 1);
        int ans = 0, s = 0;
        for (int num : nums) {
            s += num;
            ans += counter.getOrDefault(s - k, 0);
            counter.put(s, counter.getOrDefault(s, 0) + 1);
        }
        return ans;
    }
}

  • // OJ: https://leetcode.com/problems/subarray-sum-equals-k/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int subarraySum(vector<int>& nums, int k) {
        int ans = 0, sum = 0;
        unordered_map<int, int> m{ {0, 1} };
        for (int n : nums) {
            sum += n;
            auto it = m.find(sum - k);
            if (it != m.end()) ans += it->second;
            m[sum]++;
        }
        return ans;
    }
};

  • class Solution:
    def subarraySum(self, nums: List[int], k: int) -> int:
        counter = Counter({0: 1})
        ans = s = 0
        for num in nums:
            s += num
            ans += counter[s - k]
            counter[s] += 1
        return ans


class Solution: # over time limit, not fast enough
    def subarraySum(self, nums: List[int], k: int) -> int:
        count = 0

        # sum value from 1 to i-th element
        sum_arr = [0] * (len(nums) + 1)

        # cached sum
        for i in range(1, len(nums) + 1):
            sum_arr[i] = sum_arr[i - 1] + nums[i - 1]

        for start in range(len(nums)):
            for end in range(start + 1, len(nums) + 1):
                if sum_arr[end] - sum_arr[start] == k:
                    count += 1

        return count


  • func subarraySum(nums []int, k int) int {
	counter := map[int]int{0: 1}
	ans, s := 0, 0
	for _, num := range nums {
		s += num
		ans += counter[s-k]
		counter[s]++
	}
	return ans
}

  • function subarraySum(nums: number[], k: number): number {
    let ans = 0,
        s = 0;
    const counter = new Map();
    counter.set(0, 1);
    for (const num of nums) {
        s += num;
        ans += counter.get(s - k) || 0;
        counter.set(s, (counter.get(s) || 0) + 1);
    }
    return ans;
}


  • use std::collections::HashMap;

impl Solution {
    pub fn subarray_sum(nums: Vec<i32>, k: i32) -> i32 {
        let mut res = 0;
        let mut sum = 0;
        let mut map = HashMap::new();
        map.insert(0, 1);
        nums.iter().for_each(|num| {
            sum += num;
            res += map.get(&(sum - k)).unwrap_or(&0);
            map.insert(sum, map.get(&sum).unwrap_or(&0) + 1);
        });
        res
    }
}

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