package com.freetymekiyan.algorithms.level.medium;
import java.util.HashMap;
import java.util.Map;
/**
* Given an array nums and a target value k, find the maximum length of a subarray that sums to k. If there isn't one,
* return 0 instead.
* <p>
* Note:
* The sum of the entire nums array is guaranteed to fit within the 32-bit signed integer range.
* <p>
* Example 1:
* Given nums = [1, -1, 5, -2, 3], k = 3,
* return 4. (because the subarray [1, -1, 5, -2] sums to 3 and is the longest)
* <p>
* Example 2:
* Given nums = [-2, -1, 2, 1], k = 1,
* return 2. (because the subarray [-1, 2] sums to 1 and is the longest)
* <p>
* Follow Up:
* Can you do it in O(n) time?
* <p>
* Company Tags: Palantir, Facebook
* Tags: Hash Table
* Similar Problems: (M) Minimum Size Subarray Sum, (E) Range Sum Query - Immutable
*/
public class MaximumSizeSubarraySumEqualsK {
/**
* Hash Table.
* For 0 <= i < j < nums.length,
* Find max(i, j) for sum[j] - sum[i-1] = k.
* Except when i = 0, sum[j] = k.
* The brute-force way is we loop through the array, keep updating sum.
* Then subtract with each previous sum to see if it's k.
* But previous sum is already calculated, and we can use a map to get it in O(1).
* Use a hash table to store sum and its earliest index.
*/
public int maxSubArrayLen(int[] nums, int k) {
Map<Integer, Integer> map = new HashMap<>();
int res = 0;
int sum = 0;
map.put(0, -1); // Sum is 0 for index -1. Make sure sum[j] - sum[i-1] always works.
for (int i = 0; i < nums.length; i++) {
sum += nums[i];
if (map.containsKey(sum - k)) { // Search previous sums.
res = Math.max(res, i - map.get(sum - k));
}
if (!map.containsKey(sum)) { // Keep only the smallest i
map.put(sum, i);
}
}
return res;
}
}