C program to count frequency of each element in an array

Given an array arr[] of size N, the task is to find the frequency of each distinct element present in the given array.

Examples:

Input: arr[] = { 1, 100000000, 3, 100000000, 3 } 
Output: { 1 : 1, 3 : 2, 100000000 : 2 } 
Explanation:
Distinct elements of the given array are { 1, 100000000, 3 } 
Frequency of 1 in the given array is 1. 
Frequency of 100000000 in the given array is 2. 
Frequency of 3 in the given array is 2. 
Therefore, the required output is { 1 : 1, 100000000 : 2, 3 : 2 }

Input: arr[] = { 100000000, 100000000, 800000000, 100000000 } 
Output: { 100000000 : 3, 800000000 : 1}

Approach: The problem can be solved using Binary search technique. Follow the steps below to solve the problem:

• Sort the array in ascending order.
• Traverse the array and for each distinct array element, find its lower_bound index and upper_bound indices using binary search and store in variables, say LB and UB respectively. Print the value of (UB – LB) as the frequency of that element.

Below is the implementation of the above approach:

C

// C program to implement
// the above approach

#include <stdio.h>
#include <stdlib.h>

// Comparator function to sort
// the array in ascending order
int cmp(const void* a,
        const void* b)
{
    return (*(int*)a - *(int*)b);
}

// Function to find the lower_bound of X
int lower_bound(int arr[], int N, int X)
{
    // Stores minimum possible
    // value of the lower_bound
    int low = 0;

    // Stores maximum possible
    // value of the lower_bound
    int high = N;

    // Calculate the upper_bound
    // of X using binary search
    while (low < high) {

        // Stores mid element
        // of low and high
        int mid = low + (high - low) / 2;

        // If X is less than
        // or equal to arr[mid]
        if (X <= arr[mid]) {

            // Find lower_bound in
            // the left subarray
            high = mid;
        }

        else {

            // Find lower_bound in
            // the right subarray
            low = mid + 1;
        }
    }

    // Return the lower_bound index
    return low;
}

// Function to find the upper_bound of X
int upper_bound(int arr[], int N, int X)
{
    // Stores minimum possible
    // value of the upper_bound
    int low = 0;

    // Stores maximum possible
    // value of the upper_bound
    int high = N;

    // Calculate the upper_bound
    // of X using binary search
    while (low < high) {

        // Stores mid element
        // of low and high
        int mid = low + (high - low) / 2;

        // If X is greater than
        // or equal  to arr[mid]
        if (X >= arr[mid]) {

            // Find upper_bound in
            // right subarray
            low = mid + 1;
        }

        // If X is less than arr[mid]
        else {

            // Find upper_bound in
            // left subarray
            high = mid;
        }
    }

    // Return the upper_bound index
    return low;
}

// Function to find the frequency
// of an element in the array
int findFreq(int arr[], int N,
             int X)
{
    // Stores upper_bound index of X
    int UB = upper_bound(arr, N, X);

    // Stores lower_bound index of X
    int LB = lower_bound(arr, N, X);

    return (UB - LB);
}

// Utility function to print the frequency
// of each distinct element of the array
void UtilFindFreqArr(int arr[], int N)
{
    // Sort the array in
    // ascending order
    qsort(arr, N,
          sizeof(int), cmp);

    // Print start bracket
    printf("{ ");

    // Traverse the array
    for (int i = 0; i < N;) {

        // Stores frequency
        // of arr[i];
        int fr = findFreq(arr, N,
                          arr[i]);

        // Print frequency of arr[i]
        printf("%d : %d",
               arr[i], fr);

        // Update i
        i++;

        // Remove duplicate elements
        // from the array
        while (i < N && arr[i] == arr[i - 1]) {

            // Update i
            i++;
        }

        // If arr[i] is not
        // the last array element
        if (i <= N - 1) {

            printf(", ");
        }
    }

    // Print end bracket
    printf(" }");
}

// Driver Code
int main()
{
    int arr[] = { 1, 100000000, 3,
                  100000000, 3 };

    int N = sizeof(arr) / sizeof(arr[0]);

    UtilFindFreqArr(arr, N);
}

Output:

{ 1 : 1, 3 : 2, 100000000 : 2 }

Time Complexity: O(N * log(N))
Auxiliary Space: O(1)