# Count subarrays with equal number of occurrences of two given elements

• Difficulty Level : Hard
• Last Updated : 05 May, 2021

Given an array and two integers say, x and y, find the number of subarrays in which the number of occurrences of x is equal to the number of occurrences of y.

Examples:

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```Input : arr[] = {1, 2, 1},
x = 1, y = 2
Output : 2
The possible sub-arrays have same equal number
of occurrences of x and y are:
1) {1, 2}, x and y have same occurrence(1).
2) {2, 1}, x and y have same occurrence(1).

Input : arr[] = {1, 2, 1},
x = 4, y = 6
Output : 6
The possible sub-arrays have same equal number of
occurrences of x and y are:
1) {1}, x and y have same occurrence(0).
2) {2}, x and y have same occurrence(0).
3) {1}, x and y have same occurrence(0).
1) {1, 2}, x and y have same occurrence(0).
2) {2, 1}, x and y have same occurrence(0).
3) {1, 2, 1}, x and y have same occurrence(0).

Input : arr[] = {1, 2, 1},
x = 1, y = 1
Output : 6
The possible sub-arrays have same equal number
of occurrences of x and y are:
1) {1}, x and y have same occurrence(1).
2) {2}, x and y have same occurrence(0).
3) {1}, x and y have same occurrence(1).
1) {1, 2}, x and y have same occurrence(1).
2) {2, 1}, x and y have same occurrence(1).
3) {1, 2, 1}, x and y have same occurrences (2).```

Brute Force Approach (Time Complexity – O(N2)):

We can simply generate all the possible sub-arrays and check for each subarray whether the number of occurrences of x is equal to that of y in that particular subarray.

## C++

 `/* C++ program to count number of sub-arrays in which``   ``number of occurrence of x is equal to that of y``    ``using brute force */``#include ``using` `namespace` `std;` `int` `sameOccurrence(``int` `arr[], ``int` `n, ``int` `x, ``int` `y)``{``    ``int` `result = 0;` `    ``// Check for each subarray for the required condition``    ``for` `(``int` `i = 0; i <= n - 1; i++) {``        ``int` `ctX = 0,  ctY = 0;``        ``for` `(``int` `j = i; j <= n - 1; j++) {``            ``if` `(arr[j] == x)``                ``ctX += 1;``            ``else` `if` `(arr[j] == y)``                ``ctY += 1;``            ``if` `(ctX == ctY)``                ``result += 1;           ``        ``}``    ``}` `    ``return` `(result);``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, 2, 2, 3, 4, 1 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``int` `x = 2, y = 3;``    ``cout << sameOccurrence(arr, n, x, y);``    ``return` `(0);``}`

## Java

 `/* Java program to count number of sub-arrays in which``number of occurrence of x is equal to that of y``    ``using brute force */``import` `java.util.*;` `class` `solution``{` `static` `int` `sameOccurrence(``int` `arr[], ``int` `n, ``int` `x, ``int` `y)``{``    ``int` `result = ``0``;` `    ``// Check for each subarray for the required condition``    ``for` `(``int` `i = ``0``; i <= n - ``1``; i++) {``        ``int` `ctX = ``0``, ctY = ``0``;``        ``for` `(``int` `j = i; j <= n - ``1``; j++) {``            ``if` `(arr[j] == x)``                ``ctX += ``1``;``            ``else` `if` `(arr[j] == y)``                ``ctY += ``1``;``            ``if` `(ctX == ctY)``                ``result += ``1``;        ``        ``}``    ``}` `    ``return` `(result);``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``int` `arr[] = { ``1``, ``2``, ``2``, ``3``, ``4``, ``1` `};``    ``int` `n = arr.length;``    ``int` `x = ``2``, y = ``3``;``    ``System.out.println(sameOccurrence(arr, n, x, y));` `}``}` `// This code is contributed by``// Sahil_shelangia`

## Python3

 `# Python 3 program to count number of``# sub-arrays in which number of occurrence``# of x is equal to that of y using brute force``def` `sameOccurrence(arr, n, x, y):``    ``result ``=` `0` `    ``# Check for each subarray for``    ``# the required condition``    ``for` `i ``in` `range``(n):``        ``ctX ``=` `0``        ``ctY ``=` `0``        ``for` `j ``in` `range``(i, n, ``1``):``            ``if` `(arr[j] ``=``=` `x):``                ``ctX ``+``=` `1``;``            ``elif` `(arr[j] ``=``=` `y):``                ``ctY ``+``=` `1``            ``if` `(ctX ``=``=` `ctY):``                ``result ``+``=` `1` `    ``return` `(result)` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``1``, ``2``, ``2``, ``3``, ``4``, ``1``]``    ``n ``=` `len``(arr)``    ``x ``=` `2``    ``y ``=` `3``    ``print``(sameOccurrence(arr, n, x, y))` `# This code is contributed by``# Surendra_Gangwar`

## C#

 `/* C# program to count number of sub-arrays in which``number of occurrence of x is equal to that of y``using brute force */``using` `System;` `class` `GFG``{` `static` `int` `sameOccurrence(``int``[] arr, ``int` `n,``                            ``int` `x, ``int` `y)``{``    ``int` `result = 0;` `    ``// Check for each subarray for``    ``// the required condition``    ``for` `(``int` `i = 0; i <= n - 1; i++)``    ``{``        ``int` `ctX = 0, ctY = 0;``        ``for` `(``int` `j = i; j <= n - 1; j++)``        ``{``            ``if` `(arr[j] == x)``                ``ctX += 1;``            ``else` `if` `(arr[j] == y)``                ``ctY += 1;``            ``if` `(ctX == ctY)``                ``result += 1;        ``        ``}``    ``}``    ``return` `(result);``}` `// Driver code``public` `static` `void` `Main()``{``    ``int``[] arr = { 1, 2, 2, 3, 4, 1 };``    ``int` `n = arr.Length;``    ``int` `x = 2, y = 3;``    ``Console.Write(sameOccurrence(arr, n, x, y));` `}``}` `// This code is contributed by Ita_c.`

## PHP

 ``

## Javascript

 ``

Output:

`7`

Time Complexity – O(N^2)
Auxiliary Space – O(1)

Efficient Approach (O(N) Time Complexity) :

In this solution, auxiliary space is O(N) and the time complexity is also O(N). We create two arrays say, countX[] and countY[], which denotes the number of occurrences of x and y, respectively, till that point in the array. Then, we evaluate another array, say diff which stores (countX[i]-countY[i]), i be the index of the array. Now, store the count of each element of array diff in a map, say m. Initialize result as m since the occurrence of 0 in diff array gives us subarray count where the required condition is followed. Now, iterate through the map and using handshake formula, update the result, since two same values in diff array indicate that the subarray contains the same number of occurrences of x and y.

```Explanation:
arr[] = {1, 2, 2, 3, 4, 1};
x = 2, y = 3;

Two arrays countX[] and countY[] are be evaluated as-
countX[] = {0, 1, 2, 2, 2, 2};
countY[] = {0, 0, 0, 1, 1, 1};

Hence, diff[] = {0, 1, 2, 1, 1, 1};
(diff[i] = countX[i]-countY[i], i be the index of array)

Now, create a map and store the count of each element of diff in it,
so, finally, we get-
m = 1, m = 4, m = 1;

Initialize result as m
i.e result = m = 1

Further, using handshake formula, updating the
result as follows-
result  = result + (1*(1-1))/2 = 1 + 0 = 1
result  = result + (4*(4-1))/2 = 1 + 6 = 7
result  = result + (1*(1-1))/2 = 7 + 0 = 7

so, the final result will be 7, required subarrays having
same number of occurrences of x and y.```

## C++

 `/* C++ program to count number of sub-arrays in which``  ``number of occurrence of x is equal to that of y using``  ``efficient approach in terms of time */``#include ``using` `namespace` `std;` `int` `sameOccurrence(``int` `arr[], ``int` `n, ``int` `x, ``int` `y)``{``    ``int` `countX[n], countY[n];` `    ``map<``int``, ``int``> m; ``// To store counts of same diffs` `    ``// Count occurrences of x and y``    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(arr[i] == x) {``            ``if` `(i != 0)``                ``countX[i] = countX[i - 1] + 1;``            ``else``                ``countX[i] = 1;``        ``} ``else` `{``            ``if` `(i != 0)``                ``countX[i] = countX[i - 1];``            ``else``                ``countX[i] = 0;``        ``}``        ``if` `(arr[i] == y) {``            ``if` `(i != 0)``                ``countY[i] = countY[i - 1] + 1;``            ``else``                ``countY[i] = 1;``        ``} ``else` `{``            ``if` `(i != 0)``                ``countY[i] = countY[i - 1];``            ``else``                ``countY[i] = 0;``        ``}``  ` `         ``// Increment count of current``         ``m[countX[i] - countY[i]]++;``    ``}` `    ``// Traverse map and commute result.``    ``int` `result = m;``    ``for` `(``auto` `it = m.begin(); it != m.end(); it++)``        ``result = result + ((it->second) * ((it->second) - 1)) / 2;``    ` `    ``return` `(result);``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, 2, 2, 3, 4, 1 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``int` `x = 2, y = 3;``    ``cout << sameOccurrence(arr, n, x, y);``    ``return` `(0);``}`

## Java

 `/* Java program to count number of sub-arrays in which``number of occurrence of x is equal to that of y using``efficient approach in terms of time */``import` `java.util.*;` `class` `GFG``{` `static` `int` `sameOccurrence(``int` `arr[], ``int` `n, ``int` `x, ``int` `y)``{``    ``int` `[]countX = ``new` `int``[n];``    ``int` `[]countY = ``new` `int``[n];` `    ``Map m = ``new` `HashMap<>();``    ` `    ``// To store counts of same diff``    ``// Count occurrences of x and y``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{``        ``if` `(arr[i] == x)``        ``{``            ``if` `(i != ``0``)``                ``countX[i] = countX[i - ``1``] + ``1``;``            ``else``                ``countX[i] = ``1``;``        ``}``        ``else``        ``{``            ``if` `(i != ``0``)``                ``countX[i] = countX[i - ``1``];``            ``else``                ``countX[i] = ``0``;``        ``}``        ``if` `(arr[i] == y)``        ``{``            ``if` `(i != ``0``)``                ``countY[i] = countY[i - ``1``] + ``1``;``            ``else``                ``countY[i] = ``1``;``        ``}``        ``else``        ``{``            ``if` `(i != ``0``)``                ``countY[i] = countY[i - ``1``];``            ``else``                ``countY[i] = ``0``;``        ``}``    ` `        ``// Increment count of current``        ``if``(m.containsKey(countX[i] - countY[i]))``        ``{``            ``m.put(countX[i] - countY[i], m.get(countX[i] - countY[i])+``1``);``        ``}``        ``else``        ``{``            ``m.put(countX[i] - countY[i], ``1``);``        ``}``    ``}` `    ``// Traverse map and commute result.``    ``int` `result = m.get(``0``);``    ``for` `(Map.Entry it : m.entrySet())``        ``result = result + ((it.getValue()) * ((it.getValue()) - ``1``)) / ``2``;``    ` `    ``return` `(result);``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``1``, ``2``, ``2``, ``3``, ``4``, ``1` `};``    ``int` `n = arr.length;``    ``int` `x = ``2``, y = ``3``;``    ``System.out.println(sameOccurrence(arr, n, x, y));``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 program to count number of``# sub-arrays in which number of occurrence``# of x is equal to that of y using efficient``# approach in terms of time */``def` `sameOccurrence( arr, n, x, y):` `    ``countX ``=` `[``0` `for` `i ``in` `range``(n)]``    ``countY ``=` `[``0` `for` `i ``in` `range``(n)]` `    ``# To store counts of same diffs``    ``m ``=` `dict``()` `    ``# Count occurrences of x and y``    ``for` `i ``in` `range``(n):``        ``if` `(arr[i] ``=``=` `x):``            ``if` `(i !``=` `0``):``                ``countX[i] ``=` `countX[i ``-` `1``] ``+` `1``            ``else``:``                ``countX[i] ``=` `1``        ``else``:``            ``if` `(i !``=` `0``):``                ``countX[i] ``=` `countX[i ``-` `1``]``            ``else``:``                ``countX[i] ``=` `0` `        ``if` `(arr[i] ``=``=` `y):``            ``if` `(i !``=` `0``):``                ``countY[i] ``=` `countY[i ``-` `1``] ``+` `1``            ``else``:``                ``countY[i] ``=` `1``        ``else``:``            ``if` `(i !``=` `0``):``                ``countY[i] ``=` `countY[i ``-` `1``]``            ``else``:``                ``countY[i] ``=` `0``        ` `        ``# Increment count of current``        ``m[countX[i] ``-` `countY[i]] ``=` `m.get(countX[i] ``-``                                         ``countY[i], ``0``) ``+` `1``    ` `    ``# Traverse map and commute result.``    ``result ``=` `m[``0``]``    ``for` `j ``in` `m:``        ``result ``+``=` `(m[j] ``*` `(m[j] ``-` `1``)) ``/``/` `2``    ` `    ``return` `result` `# Driver code``arr ``=` `[``1``, ``2``, ``2``, ``3``, ``4``, ``1``]``n ``=` `len``(arr)``x, y ``=` `2``, ``3``print``(sameOccurrence(arr, n, x, y))`` ` `# This code is contributed``# by mohit kumar`

## C#

 `/* C# program to count number of sub-arrays in which``number of occurrence of x is equal to that of y using``efficient approach in terms of time */``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `static` `int` `sameOccurrence(``int` `[]arr, ``int` `n, ``int` `x, ``int` `y)``{``    ``int` `[]countX = ``new` `int``[n];``    ``int` `[]countY = ``new` `int``[n];` `    ``Dictionary<``int``,``int``> m = ``new` `Dictionary<``int``,``int``>();``    ` `    ``// To store counts of same diff``    ``// Count occurrences of x and y``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``if` `(arr[i] == x)``        ``{``            ``if` `(i != 0)``                ``countX[i] = countX[i - 1] + 1;``            ``else``                ``countX[i] = 1;``        ``}``        ``else``        ``{``            ``if` `(i != 0)``                ``countX[i] = countX[i - 1];``            ``else``                ``countX[i] = 0;``        ``}``        ``if` `(arr[i] == y)``        ``{``            ``if` `(i != 0)``                ``countY[i] = countY[i - 1] + 1;``            ``else``                ``countY[i] = 1;``        ``}``        ``else``        ``{``            ``if` `(i != 0)``                ``countY[i] = countY[i - 1];``            ``else``                ``countY[i] = 0;``        ``}``    ` `        ``// Increment count of current``        ``if``(m.ContainsKey(countX[i] - countY[i]))``        ``{``            ``var` `v = m[countX[i] - countY[i]]+1;``            ``m.Remove(countX[i] - countY[i]);``            ``m.Add(countX[i] - countY[i], v);``        ``}``        ``else``        ``{``            ``m.Add(countX[i] - countY[i], 1);``        ``}``    ``}` `    ``// Traverse map and commute result.``    ``int` `result = m;``    ``foreach``(KeyValuePair<``int``, ``int``> it ``in` `m)``        ``result = result + ((it.Value) * ((it.Value) - 1)) / 2;``    ` `    ``return` `(result);``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]arr = { 1, 2, 2, 3, 4, 1 };``    ``int` `n = arr.Length;``    ``int` `x = 2, y = 3;``    ``Console.WriteLine(sameOccurrence(arr, n, x, y));``}``}` `// This code is contributed by Princi Singh`

## Javascript

 ``

Output:

`7`

Time Complexity – O(N)
Auxiliary Space – O(N)

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