Given an integer array nums and an integer k, return true if nums has a continuous subarray of size at least two whose elements sum up to a multiple of k, or false otherwise.
An integer x is a multiple of k if there exists an integer n such that x = n * k. 0 is always a multiple of k.
Example 1:
Input: nums = [23,2,4,6,7], k = 6
Output: true
Explanation: [2, 4] is a continuous subarray of size 2 whose elements sum up to 6.
Example 2:
Input: nums = [23,2,6,4,7], k = 6
Output: true
Explanation: [23, 2, 6, 4, 7] is an continuous subarray of size 5 whose elements sum up to 42.
42 is a multiple of 6 because 42 = 7 * 6 and 7 is an integer.
Example 3:
Input: nums = [23,2,6,4,7], k = 13
Output: false
Constraints:
- 1 <= nums.length <= 105
- 0 <= nums[i] <= 109
- 0 <= sum(nums[i]) <= 231 - 1
- 1 <= k <= 231 - 1
Solution
class Solution {
public boolean checkSubarraySum(int[] nums, int k) {
Map<Integer, Integer> prefix = new HashMap<>();
prefix.put(0, -1);
int sum = 0;
for (int i = 0; i < nums.length; i++) {
sum += nums[i];
sum = sum % k;
Integer prev = prefix.get(sum);
if (prev != null) {
if ((i - prev) > 1) return true;
} else {
prefix.put(sum, i);
}
}
return false;
}
/*
[23, 2, 4,6,7], k = 6
[23,25,29,35,42]
[5, 1, 5, 5, 0]
5, 1, 5,
^
5: 0
1: 1
[0-2]
*/
}
Solution 2
class Solution {
public boolean checkSubarraySum(int[] nums, int k) {
int n = nums.length;
Map<Integer, Integer> map = new HashMap<>();
int[] prefix = new int[n];
map.put(0, 0);
int sum = 0;
for (int i = 0; i < prefix.length; i++) {
sum += nums[i];
prefix[i] = sum;
prefix[i] %= k;
Integer prev = map.get(prefix[i]);
if (prev != null) {
if ((i - prev) > 1) return true;
} else {
map.put(prefix[i], i);
}
}
return false;
}
}
Solution 2021-01-26
class Solution {
public boolean checkSubarraySum(int[] nums, int k) {
int n = nums.length;
Map<Integer, Integer> prefix = new HashMap<>();
prefix.put(0, -1);
int sum = 0;
for (int i = 0; i < n; i++) {
sum += nums[i];
sum = sum % k;
Integer prev = prefix.get(sum);
if (prev != null) {
if (i - prev > 1) {
return true;
}
} else {
prefix.put(sum, i);
}
}
return false;
}
}