Smallest element in an array that is repeated exactly ‘k’ times.

Given an array of size n, the goal is to find out the smallest number that is repeated exactly ‘k’ times where k > 0?
Assume that array has only positive integers and 1 <= arr[i] < 1000 for each i = 0 to n -1.

Examples:

```Input : arr[] = {2 2 1 3 1}
k = 2
Output: 1
Explanation:
Here in array,
2 is repeated 2 times
1 is repeated 2 times
3 is repeated 1 time
Hence 2 and 1 both are repeated 'k' times
i.e 2 and min(2, 1) is 1

Input : arr[] = {3 5 3 2}
k = 1
Output : 2
Explanation:
Both 2 and 5 are repeating 1 time but
min(5, 2) is 2
```

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Simple Approach: A simple approach is to use two nested loops.The outer loop picks an element one by one starting from the leftmost element. The inner loop checks if the same element is present on right side of it. If present increase the count and make the number negative which we got at the right side to prevent it from counting again.

 `// C++ program to find smallest number ` `// in array that is repeated exactly ` `// 'k' times. ` `#include ` `using` `namespace` `std; ` ` `  `const` `int` `MAX = 1000; ` ` `  `int` `findDuplicate(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``// Since arr[] has numbers in range from ` `    ``// 1 to MAX ` `    ``int` `res = MAX + 1; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(arr[i] > 0) { ` ` `  `            ``// set count to 1 as number is present ` `            ``// once ` `            ``int` `count = 1; ` `            ``for` `(``int` `j = i + 1; j < n; j++) ` `                ``if` `(arr[i] == arr[j]) ` `                    ``count += 1; ` ` `  `            ``// If frequency of number is equal to 'k' ` `            ``if` `(count == k) ` `                ``res = min(res, arr[i]); ` `        ``} ` `    ``} ` `    ``return` `res; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 2, 2, 1, 3, 1 }; ` `    ``int` `k = 2; ` `    ``int` `n = ``sizeof``(arr) / (``sizeof``(arr[0])); ` `    ``cout << findDuplicate(arr, n, k); ` `    ``return` `0; ` `} `

 `// Java program to find smallest number ` `// in array that is repeated exactly ` `// 'k' times. ` `public` `class` `GFG { ` `    ``static` `final` `int` `MAX = ``1000``; ` ` `  `    ``// finds the smallest number in arr[] ` `    ``// that is repeated k times ` `    ``static` `int` `findDuplicate(``int` `arr[], ``int` `n, ``int` `k) ` `    ``{ ` `        ``// Since arr[] has numbers in range from ` `        ``// 1 to MAX ` `        ``int` `res = MAX + ``1``; ` `        ``for` `(``int` `i = ``0``; i < n; i++) { ` `            ``if` `(arr[i] > ``0``) { ` ` `  `                ``// set count to 1 as number is ` `                ``// present once ` `                ``int` `count = ``1``; ` `                ``for` `(``int` `j = i + ``1``; j < n; j++) ` `                    ``if` `(arr[i] == arr[j]) ` `                        ``count += ``1``; ` ` `  `                ``// If frequency of number is equal ` `                ``// to 'k' ` `                ``if` `(count == k) ` `                    ``res = Math.min(res, arr[i]); ` `            ``} ` `        ``} ` `        ``return` `res; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `arr[] = { ``2``, ``2``, ``1``, ``3``, ``1` `}; ` `        ``int` `k = ``2``; ` `        ``int` `n = arr.length; ` `        ``System.out.println(findDuplicate(arr, n, k)); ` `    ``} ` `} ` `// This article is contributed by Sumit Ghosh `

 `# Python 3 program to find smallest  ` `# number in array that is repeated  ` `# exactly 'k' times. ` `MAX` `=` `1000` ` `  `def` `findDuplicate(arr, n, k): ` ` `  `    ``# Since arr[] has numbers in  ` `    ``# range from 1 to MAX ` `    ``res ``=` `MAX` `+` `1` `     `  `    ``for` `i ``in` `range``(``0``, n):  ` `        ``if` `(arr[i] > ``0``): ` ` `  `            ``# set count to 1 as number  ` `            ``# is present once ` `            ``count ``=` `1` `            ``for` `j ``in` `range``(i ``+` `1``, n): ` `                ``if` `(arr[i] ``=``=` `arr[j]): ` `                    ``count ``+``=` `1` ` `  `            ``# If frequency of number is equal to 'k' ` `            ``if` `(count ``=``=` `k): ` `                ``res ``=` `min``(res, arr[i]) ` `     `  `    ``return` `res ` ` `  `# Driver code ` `arr ``=` `[``2``, ``2``, ``1``, ``3``, ``1``]  ` `k ``=` `2` `n ``=` `len``(arr) ` `print``(findDuplicate(arr, n, k)) ` ` `  `# This code is contributed by Smitha Dinesh Semwal.  `

 `// C# program to find smallest number ` `// in array that is repeated exactly ` `// 'k' times. ` `using` `System; ` ` `  `public` `class` `GFG { ` `     `  `    ``static` `int` `MAX = 1000; ` ` `  `    ``// finds the smallest number in arr[] ` `    ``// that is repeated k times ` `    ``static` `int` `findDuplicate(``int``[] arr, ` `                              ``int` `n, ``int` `k) ` `    ``{ ` `         `  `        ``// Since arr[] has numbers in range ` `        ``// from 1 to MAX ` `        ``int` `res = MAX + 1; ` `         `  `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{ ` `            ``if` `(arr[i] > 0) ` `            ``{ ` ` `  `                ``// set count to 1 as number ` `                ``// is present once ` `                ``int` `count = 1; ` `                ``for` `(``int` `j = i + 1; j < n; j++) ` `                    ``if` `(arr[i] == arr[j]) ` `                        ``count += 1; ` ` `  `                ``// If frequency of number is ` `                ``// equal to 'k' ` `                ``if` `(count == k) ` `                    ``res = Math.Min(res, arr[i]); ` `            ``} ` `        ``} ` `         `  `        ``return` `res; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int``[] arr = { 2, 2, 1, 3, 1 }; ` `        ``int` `k = 2; ` `        ``int` `n = arr.Length; ` `         `  `        ``Console.WriteLine( ` `                      ``findDuplicate(arr, n, k)); ` `    ``} ` `} ` ` `  `// This article is contributed by vt_m. `

 ` 0)  ` `        ``{ ` ` `  `            ``// set count to 1 as number is  ` `            ``// present once ` `            ``\$count` `= 1; ` `            ``for` `(``\$j` `= ``\$i` `+ 1; ``\$j` `< ``\$n``; ``\$j``++) ` `                ``if` `(``\$arr``[``\$i``] == ``\$arr``[``\$j``]) ` `                    ``\$count` `+= 1; ` ` `  `            ``// If frequency of number is  ` `            ``// equal to 'k' ` `            ``if` `(``\$count` `== ``\$k``) ` `                ``\$res` `= min(``\$res``, ``\$arr``[``\$i``]); ` `        ``} ` `    ``} ` `    ``return` `\$res``; ` `} ` ` `  `// Driver code ` `\$arr` `= ``array``(2, 2, 1, 3, 1); ` `\$k` `= 2; ` `\$n` `= ``count``(``\$arr``); ` `echo` `findDuplicate(``\$arr``, ``\$n``, ``\$k``); ` ` `  `// This code is contributed by Rajput-Ji ` `?> `

Output:
```1
```

Time Complexity : O(n2)
Auxiliary Space : O(1)
This solution doesn’t require array elements to be in limited range.

Better Solution : Sort the input array and find the first element with exactly k count of appearances.

 `// C++ program to find smallest number ` `// in array that is repeated exactly ` `// 'k' times. ` `#include ` `using` `namespace` `std; ` ` `  `int` `findDuplicate(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``// Sort the array ` `    ``sort(arr, arr + n); ` ` `  `    ``// Find the first element with exactly ` `    ``// k occurrences. ` `    ``int` `i = 0; ` `    ``while` `(i < n) { ` `        ``int` `j, count = 1; ` `        ``for` `(j = i + 1; j < n && arr[j] == arr[i]; j++) ` `            ``count++; ` ` `  `        ``if` `(count == k) ` `            ``return` `arr[i]; ` ` `  `        ``i = j; ` `    ``} ` ` `  `    ``return` `-1; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 2, 2, 1, 3, 1 }; ` `    ``int` `k = 2; ` `    ``int` `n = ``sizeof``(arr) / (``sizeof``(arr[0])); ` `    ``cout << findDuplicate(arr, n, k); ` `    ``return` `0; ` `} `

 `// Java program to find smallest number ` `// in array that is repeated exactly ` `// 'k' times. ` `import` `java.util.Arrays; ` `public` `class` `GFG { ` ` `  `    ``// finds the smallest number in arr[] ` `    ``// that is repeated k times ` `    ``static` `int` `findDuplicate(``int` `arr[], ``int` `n, ``int` `k) ` `    ``{ ` `        ``// Sort the array ` `        ``Arrays.sort(arr); ` ` `  `        ``// Find the first element with exactly ` `        ``// k occurrences. ` `        ``int` `i = ``0``; ` `        ``while` `(i < n) { ` `            ``int` `j, count = ``1``; ` `            ``for` `(j = i + ``1``; j < n && arr[j] == arr[i]; j++) ` `                ``count++; ` ` `  `            ``if` `(count == k) ` `                ``return` `arr[i]; ` ` `  `            ``i = j; ` `        ``} ` ` `  `        ``return` `-``1``; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `arr[] = { ``2``, ``2``, ``1``, ``3``, ``1` `}; ` `        ``int` `k = ``2``; ` `        ``int` `n = arr.length; ` `        ``System.out.println(findDuplicate(arr, n, k)); ` `    ``} ` `} ` `// This article is contributed by Sumit Ghosh `

 `# Python program to find smallest number ` `# in array that is repeated exactly ` `# 'k' times. ` ` `  `def` `findDuplicate(arr, n, k): ` `     `  `    ``# Sort the array ` `    ``arr.sort() ` `  `  `    ``# Find the first element with exactly ` `    ``# k occurrences. ` `    ``i ``=` `0` `    ``while` `(i < n): ` `        ``j, count ``=` `i ``+` `1``, ``1` `        ``while` `(j < n ``and` `arr[j] ``=``=` `arr[i]): ` `            ``count ``+``=` `1` `            ``j ``+``=` `1` `  `  `        ``if` `(count ``=``=` `k): ` `            ``return` `arr[i] ` `  `  `        ``i ``=` `j ` `         `  `    ``return` `-``1` `  `  `# Driver code ` `arr ``=` `[ ``2``, ``2``, ``1``, ``3``, ``1` `]; ` `k ``=` `2` `n ``=` `len``(arr) ` `print` `findDuplicate(arr, n, k) ` ` `  `# This code is contributed by Sachin Bisht `

 `// C# program to find smallest number ` `// in array that is repeated exactly ` `// 'k' times. ` `using` `System; ` ` `  `public` `class` `GFG { ` ` `  `    ``// finds the smallest number in arr[] ` `    ``// that is repeated k times ` `    ``static` `int` `findDuplicate(``int``[] arr, ` `                             ``int` `n, ``int` `k) ` `    ``{ ` `         `  `        ``// Sort the array ` `        ``Array.Sort(arr); ` ` `  `        ``// Find the first element with ` `        ``// exactly k occurrences. ` `        ``int` `i = 0; ` `        ``while` `(i < n) { ` `            ``int` `j, count = 1; ` `            ``for` `(j = i + 1; j < n && ` `                     ``arr[j] == arr[i]; j++) ` `                ``count++; ` ` `  `            ``if` `(count == k) ` `                ``return` `arr[i]; ` ` `  `            ``i = j; ` `        ``} ` ` `  `        ``return` `-1; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int``[] arr = { 2, 2, 1, 3, 1 }; ` `        ``int` `k = 2; ` `        ``int` `n = arr.Length; ` `         `  `        ``Console.WriteLine( ` `               ``findDuplicate(arr, n, k)); ` `    ``} ` `} ` ` `  `// This article is contributed by vt_m. `

 ` `

Output:
```1
```

Time Complexity : O(n Log n)
Auxiliary Space : O(1)

Efficient Approach : Efficient approach is based on the fact that array has numbers in small range (1 to 1000). We solve this problem by using a frequency array of size max and store the frequency of every number in that array.

 `// C++ program to find smallest number ` `// in array that is repeated exactly ` `// 'k' times. ` `#include ` `using` `namespace` `std; ` ` `  `const` `int` `MAX = 1000; ` ` `  `int` `findDuplicate(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``// Computing frequencies of all elements ` `    ``int` `freq[MAX]; ` `    ``memset``(freq, 0, ``sizeof``(freq)); ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(arr[i] < 1 && arr[i] > MAX) { ` `            ``cout << ``"Out of range"``; ` `            ``return` `-1; ` `        ``} ` `        ``freq[arr[i]] += 1; ` `    ``} ` ` `  `    ``// Finding the smallest element with ` `    ``// frequency as k ` `    ``for` `(``int` `i = 0; i < MAX; i++) { ` ` `  `        ``// If frequency of any of the number ` `        ``// is equal to k starting from 0 ` `        ``// then return the number ` `        ``if` `(freq[i] == k) ` `            ``return` `i; ` `    ``} ` ` `  `    ``return` `-1; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 2, 2, 1, 3, 1 }; ` `    ``int` `k = 2; ` `    ``int` `n = ``sizeof``(arr) / (``sizeof``(arr[0])); ` `    ``cout << findDuplicate(arr, n, k); ` `    ``return` `0; ` `} `

 `// Java program to find smallest number ` `// in array that is repeated exactly ` `// 'k' times. ` `public` `class` `GFG { ` ` `  `    ``static` `final` `int` `MAX = ``1000``; ` ` `  `    ``// finds the smallest number in arr[] ` `    ``// that is repeated k times ` `    ``static` `int` `findDuplicate(``int` `arr[], ``int` `n, ``int` `k) ` `    ``{ ` `        ``// Computing frequencies of all elements ` `        ``int``[] freq = ``new` `int``[MAX]; ` ` `  `        ``for` `(``int` `i = ``0``; i < n; i++) { ` `            ``if` `(arr[i] < ``1` `&& arr[i] > MAX) { ` `                ``System.out.println(``"Out of range"``); ` `                ``return` `-``1``; ` `            ``} ` `            ``freq[arr[i]] += ``1``; ` `        ``} ` ` `  `        ``// Finding the smallest element with ` `        ``// frequency as k ` `        ``for` `(``int` `i = ``0``; i < MAX; i++) { ` ` `  `            ``// If frequency of any of the number ` `            ``// is equal to k starting from 0 ` `            ``// then return the number ` `            ``if` `(freq[i] == k) ` `                ``return` `i; ` `        ``} ` ` `  `        ``return` `-``1``; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `arr[] = { ``2``, ``2``, ``1``, ``3``, ``1` `}; ` `        ``int` `k = ``2``; ` `        ``int` `n = arr.length; ` `        ``System.out.println(findDuplicate(arr, n, k)); ` `    ``} ` `} ` `// This article is contributed by Sumit Ghosh `

 `# Python program to find smallest number ` `# in array that is repeated exactly ` `# 'k' times. ` ` `  `MAX` `=` `1000` `  `  `def` `findDuplicate(arr, n, k): ` `     `  `    ``# Computing frequencies of all elements ` `    ``freq ``=` `[``0` `for` `i ``in` `range``(``MAX``)] ` `     `  `    ``for` `i ``in` `range``(n): ` `        ``if` `(arr[i] < ``1` `and` `arr[i] > ``MAX``): ` `            ``print` `"Out of range"` `            ``return` `-``1` `        ``freq[arr[i]] ``+``=` `1` `  `  `    ``# Finding the smallest element with ` `    ``# frequency as k ` `    ``for` `i ``in` `range``(``MAX``): ` `     `  `        ``# If frequency of any of the number ` `        ``# is equal to k starting from 0 ` `        ``# then return the number ` `        ``if` `(freq[i] ``=``=` `k): ` `            ``return` `i ` `     `  `    ``return` `-``1` `     `  `# Driver code ` `arr ``=` `[ ``2``, ``2``, ``1``, ``3``, ``1` `] ` `k ``=` `2` `n ``=` `len``(arr) ` `print` `findDuplicate(arr, n, k) ` ` `  `# This code is contributed by Sachin Bisht `

 `// C# program to find smallest number ` `// in array that is repeated exactly ` `// 'k' times. ` `using` `System; ` ` `  `public` `class` `GFG { ` ` `  `    ``static` `int` `MAX = 1000; ` ` `  `    ``// finds the smallest number in arr[] ` `    ``// that is repeated k times ` `    ``static` `int` `findDuplicate(``int``[] arr, ` `                            ``int` `n, ``int` `k) ` `    ``{ ` `         `  `        ``// Computing frequencies of all ` `        ``// elements ` `        ``int``[] freq = ``new` `int``[MAX]; ` ` `  `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` `            ``if` `(arr[i] < 1 && arr[i] > MAX) ` `            ``{ ` `                ``Console.WriteLine(``"Out of range"``); ` `                ``return` `-1; ` `            ``} ` `             `  `            ``freq[arr[i]] += 1; ` `        ``} ` ` `  `        ``// Finding the smallest element with ` `        ``// frequency as k ` `        ``for` `(``int` `i = 0; i < MAX; i++) { ` ` `  `            ``// If frequency of any of the ` `            ``// number is equal to k starting ` `            ``// from 0 then return the number ` `            ``if` `(freq[i] == k) ` `                ``return` `i; ` `        ``} ` ` `  `        ``return` `-1; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int``[] arr = { 2, 2, 1, 3, 1 }; ` `        ``int` `k = 2; ` `        ``int` `n = arr.Length; ` `         `  `        ``Console.WriteLine( ` `               ``findDuplicate(arr, n, k)); ` `    ``} ` `} ` ` `  `// This article is contributed by vt_m. `

 ` ``\$MAX``) ` `        ``{ ` `            ``echo` `"Out of range"``; ` `            ``return` `-1; ` `        ``} ` `        ``\$freq``[``\$arr``[``\$i``]] += 1; ` `    ``} ` ` `  `    ``// Finding the smallest element with ` `    ``// frequency as k ` `    ``for` `(``\$i` `= 0; ``\$i` `< ``\$MAX``; ``\$i``++) ` `    ``{ ` ` `  `        ``// If frequency of any of the number ` `        ``// is equal to k starting from 0 ` `        ``// then return the number ` `        ``if` `(``\$freq``[``\$i``] == ``\$k``) ` `            ``return` `\$i``; ` `    ``} ` ` `  `    ``return` `-1; ` `} ` ` `  `    ``// Driver code ` `    ``\$arr` `= ``array``( 2, 2, 1, 3, 1 ); ` `    ``\$k` `= 2; ` `    ``\$n` `= ``count``(``\$arr``); ` `    ``echo` `findDuplicate(``\$arr``, ``\$n``, ``\$k``); ` ` `  `// This code is contributed by mits ` `?> `

Output:
```1
```

Time Complexity: O(MAX + n)
Auxiliary Space : O(MAX)

Can we solve it in O(n) time if range is not limited?
Please see Smallest element repeated exactly ‘k’ times (not limited to small range)

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