/
ContinuousSubarraySum.java
78 lines (64 loc) · 2.3 KB
/
ContinuousSubarraySum.java
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// https://leetcode.com/problems/continuous-subarray-sum
class Solution {
public boolean checkSubarraySum(int[] nums, int k) {
boolean hasContinuousZero = false;
for (int i = 1; i < nums.length; i++) {
// Prefix sum
nums[i] += nums[i - 1];
// To check for consecutive zeroes in a prefix sum
// For i = 1, it is self explanatory
// For i > 1, let arr = [33, 0, 0], prefix = [33, 33, 33]
// So in running prefix sum, in order to detect 2 consectutive zeroes
// we check nums[i] and nums[i - 2] if they are equal or not
if ((i == 1 && nums[i] == 0 && nums[i - 1] == 0) || (i > 1 && nums[i] == nums[i - 2])) {
hasContinuousZero = true;
}
}
// If we have continuous zeroes, we have a subarray
// 0 is multiple of k
if (hasContinuousZero) {
return true;
}
if (nums.length < 2) {
return false;
}
if (nums[1] % k == 0) {
return true;
}
if (nums[nums.length - 1] % k == 0) {
return true;
}
if (k > nums[nums.length - 1]) {
return hasContinuousZero;
}
NumTracker tracker = new NumTracker(k);
for (int i = 2; i < nums.length; i++) {
// Track i - 2 element so that the sub array we can form is of size atleast 2
tracker.add(nums[i - 2]);
// If prefix sum is divisible by k, we have a solution
if (nums[i] % k == 0) {
return true;
}
// If we have a target mod value which we have already tracked,
// we have a solution.
if (tracker.hasAny(nums[i])) {
return true;
}
}
return false;
}
class NumTracker {
int k;
Map<Integer, Boolean> modMap = new HashMap<>();
NumTracker(int k) {
this.k = k;
}
void add(int n) {
modMap.put(n % k, true);
}
boolean hasAny(int n) {
int target = n % k;
return modMap.getOrDefault(target, false);
}
}
}