Merge pull request #223 from arctistic/master

Author: abranhe

data-structures/Find the Unique Element

@@ -0,0 +1,32 @@
+/*
+Given an integer array of size 2N + 1. In this given array, 
+N numbers are present twice and one number is present only once in the array.
+You need to find and return that number which is unique in the array.
+*/
+
+#include<iostream>
+#include <algorithm>
+#include "solution.h"
+using namespace std;
+
+int FindUnique(int arr[], int size){
+ int xor1 = 0;
+ for(int i = 0; i < size; i++)
+ xor1 ^= arr[i];
+ return xor1;
+}
+
+int main() {
+
+int size;
+
+cin>>size;
+int *input=new int[1+size];
+
+for(int i=0;i<size;i++)
+cin>>input[i];
+
+cout<<FindUnique(input,size)<<endl;
+
+return 0;
+}

data-structures/hashmaps/longest-continious-subset-to-zero

@@ -0,0 +1,51 @@
+/*
+Given an array consisting of positive and negative integers, find the length of the longest continuous subset whose sum is zero.
+To be done in O(n) time complexity.
+
+Eg: Array size: 10 
+ Array: 95 -97 -387 -435 -5 -70 897 127 23 284
+ Then the length longest continious subset/sequence which sums up to zero is 5
+ Since the longest continious subset which sums up to zero is -387 -435 -5 -70 897.
+
+This program makes use of hashmaps and the STL is unordered_map.
+*/
+#include<iostream>
+#include <unordered_map>
+using namespace std;
+
+int lengthOfLongestSubsetWithZeroSum(int* arr, int size){
+ unordered_map<int, int> mymap;
+ int sum = 0;
+ int maxLength = -1;
+ for(int i = 0; i < size; i++){
+ sum += arr[i];
+ int length = 0;
+ 
+ if(sum == 0){
+ length = i+1;
+ }else if(mymap.count(sum)){
+ length = i - mymap[sum];
+
+ }else{
+ mymap[sum] = i;
+ }
+
+ if(length > maxLength){
+ maxLength = length;
+ }
+ }
+ return maxLength;
+}
+
+int main(){
+ int size;
+ 
+ cin >> size;
+ int* arr = new int[size];
+ for(int i = 0; i < size; i++){
+ cin >> arr[i];
+ }
+ int ans = lengthOfLongestSubsetWithZeroSum(arr,size);
+ cout << ans << endl;
+ delete arr;
+}