From a6ae0bf22de6663144bd84794e5dade858a5bc78 Mon Sep 17 00:00:00 2001
From: vishesh kedia <visheshkedia01.vk@gmail.com>
Date: Sat, 31 Aug 2019 18:53:02 +0530
Subject: [PATCH] Create Duplicate In Array

A problem solution not found in Arrays
---
 Arrays/Duplicate In Array | 37 +++++++++++++++++++++++++++++++++++++
 1 file changed, 37 insertions(+)
 create mode 100644 Arrays/Duplicate In Array

diff --git a/Arrays/Duplicate In Array b/Arrays/Duplicate In Array
new file mode 100644
index 0000000..1985fe4
--- /dev/null
+++ b/Arrays/Duplicate In Array	
@@ -0,0 +1,37 @@
+Given a read only array of n + 1 integers between 1 and n, find one number that repeats in linear time using less than O(n) space and traversing the stream sequentially O(1) times.
+
+Sample Input:
+
+[3 4 1 4 1]
+Sample Output:
+
+1
+If there are multiple possible answers ( like in the sample case above ), output any one.
+
+If there is no duplicate, output -1
+
+
+
+
+SOLUTION:
+
+int Solution::repeatedNumber(const vector<int> &A) {
+    // Do not write main() function.
+    // Do not read input, instead use the arguments to the function.
+    // Do not print the output, instead return values as specified
+    // Still have a doubt. Checkout www.interviewbit.com/pages/sample_codes/ for more details
+
+   
+    vector <bool> v(A.size());
+    fill(v.begin(), v.end(), true);
+
+    for(int i=0;i<A.size();i++)
+        {
+            if(v[A[i]])
+            {
+              v[A[i]]=false;
+            }
+            else
+             return A[i];
+    }
+}