Maximum Size Subarray Sum Equals k : Accepted

Author: gouthampradhan

README.md

@@ -128,6 +128,7 @@ My accepted leetcode solutions to some of the common interview problems.
 - [Sort Character by Frequency](problems/src/hashing/SortCharByFrequency.java) (Medium)
 - [Two Sum](problems/src/hashing/TwoSum.java) (Easy)
 - [Valid Anagram](problems/src/hashing/ValidAnagram.java) (Easy)
+- [Maximum Size Subarray Sum Equals k](problems/src/hashing/MaximumSizeSubarraySumEqualsk.java) (Medium)
 
 #### [Heap](problems/src/heap)
 

problems/src/hashing/MaximumSizeSubarraySumEqualsk.java

@@ -0,0 +1,64 @@
+package hashing;
+
+import java.util.HashMap;
+import java.util.Map;
+
+/**
+ * Created by gouthamvidyapradhan on 18/10/2017.
+ * 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.
+
+ Note:
+ The sum of the entire nums array is guaranteed to fit within the 32-bit signed integer range.
+
+ 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)
+
+ Example 2:
+ Given nums = [-2, -1, 2, 1], k = 1,
+ return 2. (because the subarray [-1, 2] sums to 1 and is the longest)
+
+ Follow Up:
+ Can you do it in O(n) time?
+ */
+public class MaximumSizeSubarraySumEqualsk {
+
+    /**
+     * Main method
+     * @param args
+     * @throws Exception
+     */
+    public static void main(String[] args) throws Exception{
+        int[] A = {1,-1,5,-2,3};
+        System.out.println(new MaximumSizeSubarraySumEqualsk().maxSubArrayLen(A, 10));
+    }
+
+    public int maxSubArrayLen(int[] nums, int k) {
+        Map<Long, Integer> index = new HashMap<>();
+        long sum = 0L;
+        for(int i = 0; i < nums.length; i ++){
+            sum += nums[i];
+            index.putIfAbsent(sum, i);
+        }
+        sum = 0;
+        int ans = 0;
+        for(int i = 0; i < nums.length; i ++){
+            sum += nums[i];
+            if(sum == k){
+                ans = Math.max(ans, i + 1);
+            } else{
+                long exp = sum - k;
+                if(index.containsKey(exp)){
+                    int farLeft = index.get(exp);
+                    if(farLeft < i){
+                        ans = Math.max(ans, i - index.get(exp));
+                    }
+                }
+            }
+        }
+
+        return ans;
+    }
+
+}