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:
- The length of the array is in range [1, 20,000].
- 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;
}
}
}
Java
class Solution {
public int subarraySum(int[] nums, int k) {
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
map.put(0, 1);
int count = 0;
int sum = 0;
int length = nums.length;
for (int i = 0; i < length; i++) {
sum += nums[i];
int curCount = map.getOrDefault(sum - k, 0);
count += curCount;
int sumCount = map.getOrDefault(sum, 0);
map.put(sum, sumCount + 1);
}
return count;
}
}