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# Mode

• Difficulty Level : Basic
• Last Updated : 02 Jul, 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 :

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

## Python3

 import numpy as npfrom 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}")
1. Mode can be useful for qualitative data.
2. Mode can be located graphically.
3. Mode can be computed in an open-end frequency table.
4. 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 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 Codeint 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 techniqueimport 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 Codeif __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 techniqueusing 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



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 the value of 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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