diff --git a/Arrays/Repeat_and_Missing_Number_Array.cpp b/Arrays/Repeat_and_Missing_Number_Array.cpp
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+++ b/Arrays/Repeat_and_Missing_Number_Array.cpp
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+/*There are certain problems which are asked in the interview to also check how you take care of overflows in your problem.
+This is one of those problems.
+Please take extra care to make sure that you are type-casting your ints to long properly and at all places. Try to verify if your solution works if number of elements is as large as 105
+
+    Food for thought :
+
+        "Even though it might not be required in this problem, in some cases, you might be required to order the operations cleverly so that the numbers do not overflow.
+        For example, if you need to calculate n! / k! where n! is factorial(n), one approach is to calculate factorial(n), factorial(k) and then divide them.
+        Another approach is to only multiple numbers from k + 1 ... n to calculate the result.
+        Obviously approach 1 is more susceptible to overflows."
+
+You are given a read only array of n integers from 1 to n.
+
+Each integer appears exactly once except A which appears twice and B which is missing.
+
+Return A and B.
+
+Note: Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
+
+Note that in your output A should precede B.
+
+Example:
+
+Input:[3 1 2 5 3] 
+
+Output:[3, 4] 
+
+A = 3, B = 4
+
+https://www.interviewbit.com/problems/repeat-and-missing-number-array/
+
+*/
+
+
+vector<int> Solution::repeatedNumber(const vector<int> &A) {
+    int n =A.size();
+    long long int sq1=0,sum1=0,sum2=0,sq2=0,x,a;
+    vector <int> v(2,0);
+    int i;
+    for(i=0;i<n;i++){
+        sum2=sum2+long(i+1);
+        sq2=sq2+long(i+1)*long(i+1);
+        sum1=sum1+long(A[i]);
+        sq1=sq1+long(A[i])*long(A[i]);
+    }
+    
+    x=(sq1-sq2)/(sum1-sum2);
+    a=(x+(sum1-sum2))/2;
+    v[0]=a;
+    v[1]=x-a;
+    return v;
+}