Mode - GeeksforGeeks

Mode

Difficulty Level : Basic
Last Updated : 02 Jun, 2021

Mode is the value which occurs most frequently in a set of observations. For example, {6, 3, 9, 6, 6, 5, 9, 3} the Mode is 6, as it occurs most often.
Fact about Mode :

Sometimes there can be more than one mode.Having two modes is called bimodal.Having more than two modes is called multimodal.
There is an empirical relationship between Mean, Median, and Mode.
Mean – Mode = 3 [ Mean – Median ]

Python3

import numpy as np
from scipy import stats
 
l1 = [12, 15, 12, 78, 54, 56, 45, 45, 18, 19, 12, 35, 67, 48, 9, 2, 45, 68]
print(f"List: {l1}")
print(f"Mean: {np.mean(l1)}")
print(f"Median: {np.median(l1)}")
print(f"Mode: {stats.mode(l1)[0]}")
 
lhs = np.mean(l1) - stats.mode(l1)[0]
rhs = 3 * (np.mean(l1) - np.median(l1))
 
print(f"LHS == RHS: {lhs == rhs}")

Mode can be useful for qualitative data.
Mode can be located graphically.
Mode can be computed in an open-end frequency table.
Mode is not affected by extremely large or small values.

Formula for Mode of grouped data :

How to find Mode ?
Naive solution:
Given an n sized unsorted array, find median and mode using counting sort technique. This can be useful when array elements are in limited range.
Examples:

Input : array a[] = {1, 1, 1, 2, 7, 1}
Output : Mode = 1
         
Input : array a[] = {9, 9, 9, 9, 9}
Output : Mode = 9

Auxiliary(count) array before summing its previous counts, c[]:
Index: 0 1 2 3 4 5 6 7 8 9 10
count: 0 4 1 0 0 0 0 1 0 0 0
Mode = index with maximum value of count.
Mode = 1(for above example)

Approach :
Assuming size of input array is n:
Step #1: Take the count array before summing its previous counts into next index.
Step #2: The index with maximum value stored in it is the mode of given data.
Step #3: In case there are more than one indexes with maximum value in it, all are results for mode so we can take any.
Step #4: Store the value at that index in a separate variable called mode.
Below is the implementation:

C++

// C++ Program for Mode using
// Counting Sort technique
#include <bits/stdc++.h>
using namespace std;
 
// Function that sort input array a[] and
// calculate mode and median using counting
// sort.
void printMode(int a[], int n)
{
    // The output array b[] will
    // have sorted array
    int b[n];
 
    // variable to store max of
    // input array which will
    // to have size of count array
    int max = *max_element(a, a + n);
 
    // auxiliary(count) array to
    // store count. Initialize
    // count array as 0. Size
    // of count array will be
    // equal to (max + 1).
    int t = max + 1;
    int count[t];
    for (int i = 0; i < t; i++)
        count[i] = 0;
 
    // Store count of each element
    // of input array
    for (int i = 0; i < n; i++)
        count[a[i]]++;
 
    // mode is the index with maximum count
    int mode = 0;
    int k = count[0];
    for (int i = 1; i < t; i++) {
        if (count[i] > k) {
            k = count[i];
            mode = i;
        }
    }
 
    cout << "mode = " << mode;
}
 
// Driver Code
int main()
{
    int a[] = { 1, 4, 1, 2, 7, 1, 2, 5, 3, 6 };
    int n = sizeof(a) / sizeof(a[0]);
    printMode(a, n);
    return 0;
}

Java

// Java Program for Mode using
// Counting Sort technique
import java.util.Arrays;
 
class GFG
{
 
    // Function that sort input array a[] and
    // calculate mode and median using counting
    // sort.
    static void printMode(int[] a, int n)
    {
        // The output array b[] will
        // have sorted array
        //int []b = new int[n];
 
        // variable to store max of
        // input array which will
        // to have size of count array
        int max = Arrays.stream(a).max().getAsInt();
 
        // auxiliary(count) array to
        // store count. Initialize
        // count array as 0. Size
        // of count array will be
        // equal to (max + 1).
        int t = max + 1;
        int[] count = new int[t];
        for (int i = 0; i < t; i++)
        {
            count[i] = 0;
        }
 
        // Store count of each element
        // of input array
        for (int i = 0; i < n; i++)
        {
            count[a[i]]++;
        }
 
        // mode is the index with maximum count
        int mode = 0;
        int k = count[0];
        for (int i = 1; i < t; i++)
        {
            if (count[i] > k)
            {
                k = count[i];
                mode = i;
            }
        }
 
        System.out.println("mode = " + mode);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] a = {1, 4, 1, 2, 7, 1, 2, 5, 3, 6};
        int n = a.length;
        printMode(a, n);
    }
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 Program for Mode using
# Counting Sort technique
 
# Function that sort input array a[] and
# calculate mode and median using counting
# sort.
def printMode(a, n) :
 
    # variable to store max of
    # input array which will
    # to have size of count array
    max_element = max(a)
 
    # auxiliary(count) array to store count.
    # Initialize count array as 0. Size
    # of count array will be equal to (max + 1).
    t = max_element + 1
    count = [0] * t
     
    for i in range(t) :
        count[i] = 0
 
    # Store count of each element
    # of input array
    for i in range(n) :
        count[a[i]] += 1
 
    # mode is the index with maximum count
    mode = 0
    k = count[0]
    for i in range(1, t) :
        if (count[i] > k) :
            k = count[i]
            mode = i
         
    print("mode = ", mode)
 
# Driver Code
if __name__ == "__main__" :
     
    a = [ 1, 4, 1, 2, 7,
          1, 2, 5, 3, 6 ]
    n = len(a)
    printMode(a, n)
 
# This code is contributed by Ryuga

C#

// C# Program for Mode using
// Counting Sort technique
using System;
using System.Linq;
public class GFG{
    
 
 
// Function that sort input array a[] and
// calculate mode and median using counting
// sort.
 static void printMode(int []a, int n)
{
    // The output array b[] will
    // have sorted array
    //int []b = new int[n];
 
    // variable to store max of
    // input array which will
    // to have size of count array
    int max =a.Max();
 
    // auxiliary(count) array to
    // store count. Initialize
    // count array as 0. Size
    // of count array will be
    // equal to (max + 1).
    int t = max + 1;
    int []count = new int[t];
    for (int i = 0; i < t; i++)
        count[i] = 0;
 
    // Store count of each element
    // of input array
    for (int i = 0; i < n; i++)
        count[a[i]]++;
 
    // mode is the index with maximum count
    int mode = 0;
    int k = count[0];
    for (int i = 1; i < t; i++) {
        if (count[i] > k) {
            k = count[i];
            mode = i;
        }
    }
 
    Console.WriteLine( "mode = " + mode);
}
 
// Driver Code
 
    static public void Main (){
    int []a = { 1, 4, 1, 2, 7, 1, 2, 5, 3, 6 };
    int n =a.Length;
    printMode(a, n);
    }
}
// This code is contributed by inder_verma

Javascript

<script>
    // Javascript Program for Mode using Counting Sort technique
     
    // Function that sort input array a[] and
    // calculate mode and median using counting
    // sort.
      function printMode(a, n)
    {
     
        // The output array b[] will
        // have sorted array
        //int []b = new int[n];
 
        // variable to store max of
        // input array which will
        // to have size of count array
        let max = Number.MIN_VALUE;
         
        for (let i = 0; i < a.length; i++)
        {
            max = Math.max(max, a[i]);
        }
 
        // auxiliary(count) array to
        // store count. Initialize
        // count array as 0. Size
        // of count array will be
        // equal to (max + 1).
        let t = max + 1;
        let count = new Array(t);
        for (let i = 0; i < t; i++)
            count[i] = 0;
 
        // Store count of each element
        // of input array
        for (let i = 0; i < n; i++)
            count[a[i]]++;
 
        // mode is the index with maximum count
        let mode = 0;
        let k = count[0];
        for (let i = 1; i < t; i++) {
            if (count[i] > k) {
                k = count[i];
                mode = i;
            }
        }
 
        document.write( "mode = " + mode);
    }
     
    let a = [ 1, 4, 1, 2, 7, 1, 2, 5, 3, 6 ];
    let n =a.length;
    printMode(a, n);
      
     // This code is contributed by surehs07.
</script>

Output:

mode = 1

Time Complexity = O(N + P), where N is the size of input array and P is size of the count array or maximum value in input array.
Auxiliary Space = O(P), where P is the size of auxiliary array.
The above solutions works good when array element values are small. Please refer below post for efficient solutions.
Most frequent element in an array
Basic Program related to Mode :

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