Create Longest subset zero sum

Author: akshayavb99

Course 2 - Data Structures in JAVA/Lecture 16 - HashMaps/Longest subset zero sum

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+/*
+Given an array consisting of positive and negative integers, find the length of the longest subarray whose sum is zero.
+
+Input Format:
+The first line of input contains an integer, that denotes the value of the size of the array. Let us denote it with the symbol N.
+The following line contains N space separated integers, that denote the value of the elements of the array.
+
+Output Format
+The first and only line of output contains length of the longest subarray whose sum is zero.
+
+Constraints:
+0 <= N <= 10^8
+Time Limit: 1 sec
+
+Sample Input 1:
+10 
+ 95 -97 -387 -435 -5 -70 897 127 23 284
+Sample Output 1:
+5
+Explanation:
+The five elements that form the longest subarray that sum up to zero are: -387, -435, -5, -70, 897 
+*/
+import java.util.HashMap;
+public class Solution {
+
+	public static int lengthOfLongestSubsetWithZeroSum(int arr[]) {
+		// Write your code here
+        HashMap<Integer,Integer> map = new HashMap<>();
+        int sum=0,maxLen=-1;
+        for (int i=0;i<arr.length;i++)
+        {
+            sum=sum+arr[i];
+            //System.out.println("Current cumulative sum: "+sum);
+            if (sum==0)
+            {
+                maxLen=i+1;
+            }
+            if (map.containsKey(sum))
+            {
+                int prevIndex=map.get(sum);
+                //System.out.println("Sum found previously at index: "+prevIndex);
+                int currLen=i-prevIndex;
+                //System.out.println("Length of 0 sum: "+currLen);
+                if (currLen>maxLen)
+                {
+                    maxLen=currLen;
+                    //System.out.println("Max Length of 0 sum: "+maxLen);
+                }
+            }
+            else
+            {
+                //System.out.println("Adding sum to HashMap");
+                map.put(sum,i);
+            }
+            //System.out.println();
+        }
+        return maxLen;
+	}
+}