Given an array of n non-negative integers. The task is to find frequency of a particular element in the arbitrary range of array[]. The range is given as positions (not 0 based indexes) in array. There can be multiple queries of given type. **Examples:**

Input : arr[] = {2, 8, 6, 9, 8, 6, 8, 2, 11}; left = 2, right = 8, element = 8 left = 2, right = 5, element = 6 Output : 3 1 The element 8 appears 3 times in arr[left-1..right-1] The element 6 appears 1 time in arr[left-1..right-1]

**Naive approach: **is to traverse from left to right and update count variable whenever we find the element.

Below is the code of Naive approach:-

## C++

`// C++ program to find total count of an element` `// in a range` `#include<bits/stdc++.h>` `using` `namespace` `std;` `// Returns count of element in arr[left-1..right-1]` `int` `findFrequency(` `int` `arr[], ` `int` `n, ` `int` `left,` ` ` `int` `right, ` `int` `element)` `{` ` ` `int` `count = 0;` ` ` `for` `(` `int` `i=left-1; i<=right; ++i)` ` ` `if` `(arr[i] == element)` ` ` `++count;` ` ` `return` `count;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `arr[] = {2, 8, 6, 9, 8, 6, 8, 2, 11};` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `// Print frequency of 2 from position 1 to 6` ` ` `cout << ` `"Frequency of 2 from 1 to 6 = "` ` ` `<< findFrequency(arr, n, 1, 6, 2) << endl;` ` ` `// Print frequency of 8 from position 4 to 9` ` ` `cout << ` `"Frequency of 8 from 4 to 9 = "` ` ` `<< findFrequency(arr, n, 4, 9, 8);` ` ` `return` `0;` `}` |

## Java

`// JAVA Code to find total count of an element` `// in a range` `class` `GFG {` ` ` ` ` `// Returns count of element in arr[left-1..right-1]` ` ` `public` `static` `int` `findFrequency(` `int` `arr[], ` `int` `n,` ` ` `int` `left, ` `int` `right,` ` ` `int` `element)` ` ` `{` ` ` `int` `count = ` `0` `;` ` ` `for` `(` `int` `i = left - ` `1` `; i < right; ++i)` ` ` `if` `(arr[i] == element)` ` ` `++count;` ` ` `return` `count;` ` ` `}` ` ` ` ` `/* Driver program to test above function */` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `arr[] = {` `2` `, ` `8` `, ` `6` `, ` `9` `, ` `8` `, ` `6` `, ` `8` `, ` `2` `, ` `11` `};` ` ` `int` `n = arr.length;` ` ` ` ` `// Print frequency of 2 from position 1 to 6` ` ` `System.out.println(` `"Frequency of 2 from 1 to 6 = "` `+` ` ` `findFrequency(arr, n, ` `1` `, ` `6` `, ` `2` `));` ` ` ` ` `// Print frequency of 8 from position 4 to 9` ` ` `System.out.println(` `"Frequency of 8 from 4 to 9 = "` `+` ` ` `findFrequency(arr, n, ` `4` `, ` `9` `, ` `8` `));` ` ` ` ` `}` ` ` `}` `// This code is contributed by Arnav Kr. Mandal.` |

## Python3

`# Python program to find total ` `# count of an element in a range` `# Returns count of element` `# in arr[left-1..right-1]` `def` `findFrequency(arr, n, left, right, element):` ` ` `count ` `=` `0` ` ` `for` `i ` `in` `range` `(left ` `-` `1` `, right):` ` ` `if` `(arr[i] ` `=` `=` `element):` ` ` `count ` `+` `=` `1` ` ` `return` `count` `# Driver Code` `arr ` `=` `[` `2` `, ` `8` `, ` `6` `, ` `9` `, ` `8` `, ` `6` `, ` `8` `, ` `2` `, ` `11` `]` `n ` `=` `len` `(arr)` `# Print frequency of 2 from position 1 to 6` `print` `(` `"Frequency of 2 from 1 to 6 = "` `,` ` ` `findFrequency(arr, n, ` `1` `, ` `6` `, ` `2` `))` `# Print frequency of 8 from position 4 to 9` `print` `(` `"Frequency of 8 from 4 to 9 = "` `,` ` ` `findFrequency(arr, n, ` `4` `, ` `9` `, ` `8` `))` ` ` ` ` `# This code is contributed by Anant Agarwal.` |

## C#

`// C# Code to find total count` `// of an element in a range` `using` `System;` `class` `GFG {` ` ` ` ` `// Returns count of element` ` ` `// in arr[left-1..right-1]` ` ` `public` `static` `int` `findFrequency(` `int` `[]arr, ` `int` `n,` ` ` `int` `left, ` `int` `right,` ` ` `int` `element)` ` ` `{` ` ` `int` `count = 0;` ` ` `for` `(` `int` `i = left - 1; i < right; ++i)` ` ` `if` `(arr[i] == element)` ` ` `++count;` ` ` `return` `count;` ` ` `}` ` ` ` ` `// Driver Code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `[]arr = {2, 8, 6, 9, 8, 6, 8, 2, 11};` ` ` `int` `n = arr.Length;` ` ` ` ` `// Print frequency of 2` ` ` `// from position 1 to 6` ` ` `Console.WriteLine(` `"Frequency of 2 from 1 to 6 = "` `+` ` ` `findFrequency(arr, n, 1, 6, 2));` ` ` ` ` `// Print frequency of 8` ` ` `// from position 4 to 9` ` ` `Console.Write(` `"Frequency of 8 from 4 to 9 = "` `+` ` ` `findFrequency(arr, n, 4, 9, 8));` ` ` ` ` `}` `}` `// This code is contributed by Nitin Mittal.` |

## PHP

`<?php` `// PHP program to find total count of` `// an element in a range` `// Returns count of element in` `// arr[left-1..right-1]` `function` `findFrequency(&` `$arr` `, ` `$n` `, ` `$left` `,` ` ` `$right` `, ` `$element` `)` `{` ` ` `$count` `= 0;` ` ` `for` `(` `$i` `= ` `$left` `- 1; ` `$i` `<= ` `$right` `; ++` `$i` `)` ` ` `if` `(` `$arr` `[` `$i` `] == ` `$element` `)` ` ` `++` `$count` `;` ` ` `return` `$count` `;` `}` `// Driver Code` `$arr` `= ` `array` `(2, 8, 6, 9, 8, 6, 8, 2, 11);` `$n` `= sizeof(` `$arr` `);` `// Print frequency of 2 from position 1 to 6` `echo` `"Frequency of 2 from 1 to 6 = "` `.` ` ` `findFrequency(` `$arr` `, ` `$n` `, 1, 6, 2) .` `"\n"` `;` `// Print frequency of 8 from position 4 to 9` `echo` `"Frequency of 8 from 4 to 9 = "` `.` ` ` `findFrequency(` `$arr` `, ` `$n` `, 4, 9, 8);` `// This code is contributed by ita_c` `?>` |

## Javascript

`<script>` `// Javascript Code to find total count of an element` `// in a range` ` ` ` ` `// Returns count of element in arr[left-1..right-1]` ` ` `function` `findFrequency(arr,n,left,right,element)` ` ` `{` ` ` `let count = 0;` ` ` `for` `(let i = left - 1; i < right; ++i)` ` ` `if` `(arr[i] == element)` ` ` `++count;` ` ` `return` `count;` ` ` `}` ` ` ` ` `/* Driver program to test above function */` ` ` `let arr=[2, 8, 6, 9, 8, 6, 8, 2, 11];` ` ` `let n = arr.length;` ` ` ` ` `// Print frequency of 2 from position 1 to 6` ` ` `document.write(` `"Frequency of 2 from 1 to 6 = "` `+` ` ` `findFrequency(arr, n, 1, 6, 2)+` `"<br>"` `);` ` ` ` ` `// Print frequency of 8 from position 4 to 9` ` ` `document.write(` `"Frequency of 8 from 4 to 9 = "` `+` ` ` `findFrequency(arr, n, 4, 9, 8));` ` ` ` ` `// This code is contributed by rag2127` ` ` `</script>` |

**Output: **

Frequency of 2 from 1 to 6 = 1 Frequency of 8 from 4 to 9 = 2

**Time complexity** of this approach is O(right – left + 1) or O(n) **Auxiliary space**: O(1)

An **Efficient approach** is to use hashing. In C++, we can use unordered_map

- At first, we will store the position in map[] of every distinct element as a vector like that

int arr[] = {2, 8, 6, 9, 8, 6, 8, 2, 11}; map[2] = {1, 8} map[8] = {2, 5, 7} map[6] = {3, 6} ans so on...

- As we can see that elements in map[] are already in sorted order (Because we inserted elements from left to right), the answer boils down to find the total count in that hash map[] using binary search like method.

- In C++ we can use lower_bound which will returns an iterator pointing to the first element in the range [first, last] which has a value not less than ‘left’. and upper_bound returns an iterator pointing to the first element in the range [first,last) which has a value greater than ‘right’.

- After that we just need to subtract the upper_bound() and lower_bound() result to get the final answer. For example, suppose if we want to find the total count of 8 in the range from [1 to 6], then the map[8] of lower_bound() function will return the result 0 (pointing to 2) and upper_bound() will return 2 (pointing to 7), so we need to subtract the both the result like 2 – 0 = 2 .

Below is the code of above approach

## C++

`// C++ program to find total count of an element` `#include<bits/stdc++.h>` `using` `namespace` `std;` `unordered_map< ` `int` `, vector<` `int` `> > store;` `// Returns frequency of element in arr[left-1..right-1]` `int` `findFrequency(` `int` `arr[], ` `int` `n, ` `int` `left,` ` ` `int` `right, ` `int` `element)` `{` ` ` `// Find the position of first occurrence of element` ` ` `int` `a = lower_bound(store[element].begin(),` ` ` `store[element].end(),` ` ` `left)` ` ` `- store[element].begin();` ` ` `// Find the position of last occurrence of element` ` ` `int` `b = upper_bound(store[element].begin(),` ` ` `store[element].end(),` ` ` `right)` ` ` `- store[element].begin();` ` ` `return` `b-a;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `arr[] = {2, 8, 6, 9, 8, 6, 8, 2, 11};` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `// Storing the indexes of an element in the map` ` ` `for` `(` `int` `i=0; i<n; ++i)` ` ` `store[arr[i]].push_back(i+1); ` `//starting index from 1` ` ` `// Print frequency of 2 from position 1 to 6` ` ` `cout << ` `"Frequency of 2 from 1 to 6 = "` ` ` `<< findFrequency(arr, n, 1, 6, 2) <<endl;` ` ` `// Print frequency of 8 from position 4 to 9` ` ` `cout << ` `"Frequency of 8 from 4 to 9 = "` ` ` `<< findFrequency(arr, n, 4, 9, 8);` ` ` `return` `0;` `}` |

## Python3

`# Python3 program to find total count of an element` `from` `collections ` `import` `defaultdict as ` `dict` `from` `bisect ` `import` `bisect_left as lower_bound` `from` `bisect ` `import` `bisect_right as upper_bound` `store ` `=` `dict` `(` `list` `)` `# Returns frequency of element` `# in arr[left-1..right-1]` `def` `findFrequency(arr, n, left, right, element):` ` ` ` ` `# Find the position of` ` ` `# first occurrence of element` ` ` `a ` `=` `lower_bound(store[element], left)` ` ` `# Find the position of` ` ` `# last occurrence of element` ` ` `b ` `=` `upper_bound(store[element], right)` ` ` `return` `b ` `-` `a` `# Driver code` `arr ` `=` `[` `2` `, ` `8` `, ` `6` `, ` `9` `, ` `8` `, ` `6` `, ` `8` `, ` `2` `, ` `11` `]` `n ` `=` `len` `(arr)` `# Storing the indexes of` `# an element in the map` `for` `i ` `in` `range` `(n):` ` ` `store[arr[i]].append(i ` `+` `1` `)` `# Prfrequency of 2 from position 1 to 6` `print` `(` `"Frequency of 2 from 1 to 6 = "` `,` ` ` `findFrequency(arr, n, ` `1` `, ` `6` `, ` `2` `))` `# Prfrequency of 8 from position 4 to 9` `print` `(` `"Frequency of 8 from 4 to 9 = "` `,` ` ` `findFrequency(arr, n, ` `4` `, ` `9` `, ` `8` `))` `# This code is contributed by Mohit Kumar` |

**Output: **

Frequency of 2 from 1 to 6 = 1 Frequency of 8 from 4 to 9 = 2

**This approach will be beneficial if we have a large number of queries of an arbitrary range asking the total frequency of particular element.****Time complexity: **O(log N) for single query.

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