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MinimumSizeSubarraySum.java
64 lines (53 loc) · 1.94 KB
/
MinimumSizeSubarraySum.java
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class Solution {
public int minSubArrayLen(int target, int[] nums) {
// [2, 5, 6, 8, 12, 15]
return slidingWindow(target, nums);
}
// Prefix sum binary search approach
// We have to discard indexes found by binary search that are greater than array bounds.
// Logic is to build the prefix sum array, and check where the potential target could be in the prefix sum array.
// Works only when all elements in nums are positive.
// 5ms
int prefixSum(int target, int[] nums) {
int[] prefix = nums.clone();
for (int i = 1; i < nums.length; i++) {
prefix[i] += prefix[i - 1];
}
int min = Integer.MAX_VALUE;
for (int i = 0; i < nums.length; i++) {
int ix;
if (i == 0) {
ix = Arrays.binarySearch(prefix, target);
} else {
ix = Arrays.binarySearch(prefix, target + prefix[i - 1]);
}
if (ix < 0) ix = -ix - 1;
if (ix >= nums.length) continue;
min = Math.min(min, ix - i + 1);
}
return min == Integer.MAX_VALUE ? 0 : min;
}
// Sliding window approach
// l and r are both inclusive
// count = r - l + 1
int slidingWindow(int target, int[] nums) {
int l = 0, r = 0, min = Integer.MAX_VALUE;
int rollingSum = nums[0];
for (; l < nums.length;) {
if (rollingSum < target) {
if (r < nums.length - 1) {
r += 1;
rollingSum += nums[r];
} else {
rollingSum -= nums[l];
l += 1;
}
} else {
min = Math.min(min, r - l + 1);
rollingSum -= nums[l];
l += 1;
}
}
return min == Integer.MAX_VALUE ? 0 : min;
}
}