diff --git a/algorithms/CPlusPlus/Arrays/repeat-and-missing-number.cpp b/algorithms/CPlusPlus/Arrays/repeat-and-missing-number.cpp
new file mode 100644
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+++ b/algorithms/CPlusPlus/Arrays/repeat-and-missing-number.cpp
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+/* Problem Statement: You are given a read-only array of N
+integers with values also in the range [1, N] both inclusive.
+Each integer appears exactly once except A which appears twice and B which is missing.
+The task is to find the repeating and missing numbers A and B where A repeats twice and B is missing.
+
+Example 1:
+
+Input Format:  array[] = {3,1,2,5,3}
+
+Result: {3,4)
+
+Explanation: A = 3 , B = 4
+Since 3 is appearing twice and 4 is missing
+
+Example 2:
+
+Input Format: array[] = {3,1,2,5,4,6,7,5}
+
+Result: {5,8)
+
+Explanation: A = 5 , B = 8
+Since 5 is appearing twice and 8 is missing
+
+Approach:
+Since the numbers are from 1 to N in the array arr[]
+Take a substitute array of size N+1 and initalize it with 0.
+Traverse the given array and increase the value of substitute[arr[i]] by one .
+Then again traverse the substitute array starting from index 1 to N.
+If you find any index value greater than 1 that is repeating element A.
+If you find any index value = 0 then that is the missing element B.
+
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+void find_dup_miss(int a[], int n)
+{
+    int miss, dup;
+    int temp[n + 1] = {0};
+    for (int i = 0; i < n; i++)
+    {
+        temp[a[i]] += 1;
+    }
+    for (int i = 1; i <= n; i++)
+    {
+        if (temp[i] == 0)
+        {
+            miss = i;
+        }
+        if (temp[i] > 1)
+        {
+            dup = i;
+        }
+    }
+    cout << "duplicate element is " << dup << " and missing number is " << miss;
+}
+
+int main()
+{
+    cout << "enter the value of n: ";
+    int n;
+    cin >> n;
+    cout << "enter the array: ";
+    int a[n];
+    for (int i = 0; i < n; i++)
+    {
+        cin >> a[i];
+    }
+    find_dup_miss(a, n);
+    return 0;
+}
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