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Daily/duplicate_in_array.cpp

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+/*
+Given an array a of size N which contains elements from 0 to N-1, you need to find all the 
+elements occurring more than once in the given array. Return the answer in ascending order. 
+If no such element is found, return list containing [-1]. 
+
+Note: The extra space is only for the array to be returned. Try and perform all operations 
+within the provided array. 
+
+Example 1:
+
+Input:
+N = 4
+a[] = {0,3,1,2}
+Output: 
+-1
+Explanation: 
+There is no repeating element in the array. Therefore output is -1.
+
+Example 2:
+
+Input:
+N = 5
+a[] = {2,3,1,2,3}
+Output: 
+2 3 
+Explanation: 
+2 and 3 occur more than once in the given array.*/
+
+//{ Driver Code Starts
+#include <bits/stdc++.h>
+using namespace std;
+
+// } Driver Code Ends
+class Solution{
+  public:
+    vector<int> duplicates(int arr[], int n) {
+        map<int,int>mp;
+        vector<int>v;
+        for(int i=0;i<n;i++)
+        {
+            mp[arr[i]]++;
+        }
+        for(auto i=mp.begin();i!=mp.end();i++)
+        {
+            if(i->second>1) v.push_back(i->first);
+        }
+
+        if(v.size()==0)return {-1};
+        return v;
+    }
+};
+
+
+//{ Driver Code Starts.
+int main() {
+    int t;
+    cin >> t;
+    while (t-- > 0) {
+        int n;
+        cin >> n;
+        int a[n];
+        for (int i = 0; i < n; i++) cin >> a[i];
+        Solution obj;
+        vector<int> ans = obj.duplicates(a, n);
+        for (int i : ans) cout << i << ' ';
+        cout << endl;
+    }
+    return 0;
+}
+
+// } Driver Code Ends