# N/3 repeated number in an array with O(1) space

We are given a read only array of n integers. Find any element that appears more than n/3 times in the array in linear time and constant additional space. If no such element exists, return -1.

Examples:

```Input : [10, 10, 20, 30, 10, 10]
Output : 10
10 occurs 4 times which is more than 6/3.

Input : [20, 30, 10, 10, 5, 4, 20, 1, 2]
Output : -1```
Recommended Practice

The idea is based on Moore’s Voting algorithm.  We first find two candidates. Then we check if any of these two candidates is actually a majority. Below is the solution for above approach.

Implementation:

## C++

 `// CPP program to find if any element appears` `// more than n/3.` `#include ` `using` `namespace` `std;`   `int` `appearsNBy3(``int` `arr[], ``int` `n)` `{` `    ``int` `count1 = 0, count2 = 0;` `    ``int` `first=INT_MAX    , second=INT_MAX    ;`   `    ``for` `(``int` `i = 0; i < n; i++) {`   `        ``// if this element is previously seen, ` `        ``// increment count1.` `        ``if` `(first == arr[i])` `            ``count1++;`   `        ``// if this element is previously seen, ` `        ``// increment count2.` `        ``else` `if` `(second == arr[i])` `            ``count2++;` `    `  `        ``else` `if` `(count1 == 0) {` `            ``count1++;` `            ``first = arr[i];` `        ``}`   `        ``else` `if` `(count2 == 0) {` `            ``count2++;` `            ``second = arr[i];` `        ``}`   `        ``// if current element is different from` `        ``// both the previously seen variables, ` `        ``// decrement both the counts.` `        ``else` `{` `            ``count1--;` `            ``count2--;` `        ``}` `    ``}`   `    ``count1 = 0;` `    ``count2 = 0;`   `    ``// Again traverse the array and find the` `    ``// actual counts.` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``if` `(arr[i] == first)` `            ``count1++;`   `        ``else` `if` `(arr[i] == second)` `            ``count2++;` `    ``}`   `    ``if` `(count1 > n / 3)` `        ``return` `first;`   `    ``if` `(count2 > n / 3)` `        ``return` `second;`   `    ``return` `-1;` `}`   `int` `main()` `{` `    ``int` `arr[] = { 1, 2, 3, 1, 1 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``cout << appearsNBy3(arr, n) << endl;` `    ``return` `0;` `}`

## Java

 `// Java program to find if any element appears` `// more than n/3.` `class` `GFG {` `    `  `    ``static` `int` `appearsNBy3(``int` `arr[], ``int` `n)` `    ``{` `        ``int` `count1 = ``0``, count2 = ``0``;` `        `  `        ``// take the integers as the maximum ` `        ``// value of integer hoping the integer` `        ``// would not be present in the array` `        ``int` `first =  Integer.MIN_VALUE;` `        ``int` `second = Integer.MAX_VALUE;` `    `  `        ``for` `(``int` `i = ``0``; i < n; i++) {` `    `  `            ``// if this element is previously` `            ``// seen, increment count1.` `            ``if` `(first == arr[i])` `                ``count1++;` `    `  `            ``// if this element is previously` `            ``// seen, increment count2.` `            ``else` `if` `(second == arr[i])` `                ``count2++;` `        `  `            ``else` `if` `(count1 == ``0``) {` `                ``count1++;` `                ``first = arr[i];` `            ``}` `    `  `            ``else` `if` `(count2 == ``0``) {` `                ``count2++;` `                ``second = arr[i];` `            ``}` `    `  `            ``// if current element is different` `            ``// from both the previously seen` `            ``// variables, decrement both the` `            ``// counts.` `            ``else` `{` `                ``count1--;` `                ``count2--;` `            ``}` `        ``}` `    `  `        ``count1 = ``0``;` `        ``count2 = ``0``;` `    `  `        ``// Again traverse the array and ` `        ``// find the actual counts.` `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``if` `(arr[i] == first)` `                ``count1++;` `    `  `            ``else` `if` `(arr[i] == second)` `                ``count2++;` `        ``}` `    `  `        ``if` `(count1 > n / ``3``)` `            ``return` `first;` `    `  `        ``if` `(count2 > n / ``3``)` `            ``return` `second;` `    `  `        ``return` `-``1``;` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``int` `arr[] = { ``1``, ``2``, ``3``, ``1``, ``1` `};` `        ``int` `n = arr.length;` `        ``System.out.println(appearsNBy3(arr, n));` `        `  `    ``}` `}`   `// This code is contributed by Arnab Kundu`

## Python 3

 `# Python 3 program to find if ` `# any element appears more than` `# n/3.` `import` `sys`   `def` `appearsNBy3(arr, n):`   `    ``count1 ``=` `0` `    ``count2 ``=` `0` `    ``first ``=` `sys.maxsize` `    ``second ``=` `sys.maxsize`   `    ``for` `i ``in` `range``(``0``, n): `   `        ``# if this element is` `        ``# previously seen, ` `        ``# increment count1.` `        ``if` `(first ``=``=` `arr[i]):` `            ``count1 ``+``=` `1`   `        ``# if this element is` `        ``# previously seen, ` `        ``# increment count2.` `        ``elif` `(second ``=``=` `arr[i]):` `            ``count2 ``+``=` `1` `    `  `        ``elif` `(count1 ``=``=` `0``):` `            ``count1 ``+``=` `1` `            ``first ``=` `arr[i]`   `        ``elif` `(count2 ``=``=` `0``):` `            ``count2 ``+``=` `1` `            ``second ``=` `arr[i]` `        `    `        ``# if current element is ` `        ``# different from both` `        ``# the previously seen ` `        ``# variables, decrement` `        ``# both the counts.` `        ``else``:` `            ``count1 ``-``=` `1` `            ``count2 ``-``=` `1` `        `  `    `    `    ``count1 ``=` `0` `    ``count2 ``=` `0`   `    ``# Again traverse the array` `    ``# and find the actual counts.` `    ``for` `i ``in` `range``(``0``, n): ` `        ``if` `(arr[i] ``=``=` `first):` `            ``count1 ``+``=` `1`   `        ``elif` `(arr[i] ``=``=` `second):` `            ``count2 ``+``=` `1` `    `    `    ``if` `(count1 > n ``/` `3``):` `        ``return` `first`   `    ``if` `(count2 > n ``/` `3``):` `        ``return` `second`   `    ``return` `-``1`   `# Driver code` `arr ``=` `[``1``, ``2``, ``3``, ``1``, ``1` `]` `n ``=` `len``(arr) ` `print``(appearsNBy3(arr, n))`   `# This code is contributed by ` `# Smitha`

## C#

 `// C# program to find if any element appears` `// more than n/3.` `using` `System;`   `public` `class` `GFG {` `    `  `    ``static` `int` `appearsNBy3(``int` `[]arr, ``int` `n)` `    ``{` `        ``int` `count1 = 0, count2 = 0;` `        `  `        ``// take the integers as the maximum` `        ``// value of integer hoping the integer` `        ``// would not be present in the array` `        ``int` `first = ``int``.MaxValue;` `        ``int` `second = ``int``.MaxValue;` `    `  `        ``for` `(``int` `i = 1; i < n; i++) {` `    `  `            ``// if this element is previously ` `            ``// seen, increment count1.` `            ``if` `(first == arr[i])` `                ``count1++;` `    `  `            ``// if this element is previously ` `            ``// seen, increment count2.` `            ``else` `if` `(second == arr[i])` `                ``count2++;` `        `  `            ``else` `if` `(count1 == 0) {` `                ``count1++;` `                ``first = arr[i];` `            ``}` `    `  `            ``else` `if` `(count2 == 0) {` `                ``count2++;` `                ``second = arr[i];` `            ``}` `    `  `            ``// if current element is different` `            ``// from both the previously seen ` `            ``// variables, decrement both the` `            ``// counts.` `            ``else` `{` `                ``count1--;` `                ``count2--;` `            ``}` `        ``}` `    `  `        ``count1 = 0;` `        ``count2 = 0;` `    `  `        ``// Again traverse the array and` `        ``// find the actual counts.` `        ``for` `(``int` `i = 0; i < n; i++) {` `            ``if` `(arr[i] == first)` `                ``count1++;` `    `  `            ``else` `if` `(arr[i] == second)` `                ``count2++;` `        ``}` `    `  `        ``if` `(count1 > n / 3)` `            ``return` `first;` `    `  `        ``if` `(count2 > n / 3)` `            ``return` `second;` `    `  `        ``return` `-1;` `    ``}` `    `  `    ``// Driver code` `    ``static` `public` `void` `Main(String []args)` `    ``{` `        ``int` `[]arr = { 1, 2, 3, 1, 1 };` `        ``int` `n = arr.Length;` `        ``Console.WriteLine(appearsNBy3(arr, n));` `    ``}` `}`   `// This code is contributed by Arnab Kundu`

## PHP

 ` ``\$n` `/ 3)` `        ``return` `\$first``;`   `    ``if` `(``\$count2` `> ``\$n` `/ 3)` `        ``return` `\$second``;`   `    ``return` `-1;` `}`   `// Driver code` `\$arr` `= ``array``( 1, 2, 3, 1, 1 );` `\$n` `= ``count``(``\$arr``);` `echo` `appearsNBy3(``\$arr``, ``\$n``) ;`   `// This code is contributed by anuj_67.` `?>`

## Javascript

 ``

Output

`1`

Complexity Analysis:

• Time Complexity:  O(n)
First pass of the algorithm takes complete traversal over the array contributing to O(n) and another O(n) is consumed in checking if count1 and count2 is greater than floor(n/3) times.
• Space Complexity: O(1)
As no extra space is required so space complexity is constant

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