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.

MaxSubArrayLen.java

/*
To find: A range i..j such that nums[i] + nums[i+1] + ... + nums[j] = sum(i, j) = k
sum(i, j) = sum(0, j) - sum(0, i-1) since for [1 2 3 4], sum(1, 3) = sum(0, 3) - sum(0, 0);
sum(0, j) - sum(0, i-1) = k,  or sum(0, i-1) = sum(0, j) - k 
For each sum(0, j), check if there was a sum(0, i-1) such that it equals sum(0, j) - k.
*/

public int maxSubArrayLen(int[] nums, int k) {
    int sum = 0, max = 0;
    Map<Integer, Integer> map = new HashMap<>();
    map.put(0, -1);                                 // incase i = 0

    for(int j = 0; j < nums.length; j++) {
        sum += nums[j];                             // sum is sum(0, j)
        if(!map.containsKey(sum)) map.put(sum, j);

        if(map.containsKey(sum - k))         
            max = Math.max(max, j - map.get(sum-k)); // i = map.get(sum-k)
    }

    return max;
}