Mode in a stream of integers (running integers)

• Difficulty Level : Medium
• Last Updated : 18 Sep, 2020

Given that integers are being read from a data stream. Find the mode of all the elements read so far starting from the first integer till the last integer.

Mode is defined as the element which occurs the maximum time. If two or more elements have the same maximum frequency, then take the one with the last occurrence.

Examples:

Input: stream[] = {2, 7, 3, 2, 5}
Output: 2 7 3 2 2
Explanation:
Mode of Running Stream is computed as follows:
Mode({2}) = 2
Mode({2, 7}) = 7
Mode({2, 7, 3}) = 3
Mode({2, 7, 3, 2}) = 2
Mode({2, 7, 3, 2, 2}) = 2

Input: stream[] = {3, 5, 9, 9, 2, 3, 3, 4}
Output: 3 5 9 9 9 3 3 3

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to use a Hash-map to map elements to its frequency. While reading the elements one by one update the frequencies of elements in the map and also update the mode which will be the mode of the stream of the running integers.

Below is the implementation of the above approach:

C++

 // C++ program to implement// the above approach#includeusing namespace std;  // Function that prints// the Mode valuesvoid findMode(int a[], int n){          // Map used to mp integers    // to its frequency    map mp;       // To store the maximum frequency    int max = 0;      // To store the element with    // the maximum frequency    int mode = 0;          // Loop used to read the    // elements one by one    for(int i = 0; i < n; i++)     {                  // Updates the frequency of        // that element        mp[a[i]]++;              // Checks for maximum Number        // of occurrence        if (mp[a[i]] >= max)         {                  // Updates the maximum frequency            max = mp[a[i]];                  // Updates the Mode            mode = a[i];        }        cout << mode << " ";    }}      // Driver Codeint main(){    int arr[] = { 2, 7, 3, 2, 5 };    int n = sizeof(arr)/sizeof(arr);      // Function call    findMode(arr, n);      return 0;}   // This code is contributed by rutvik_56

Java

 // Java implementation of the// above approach  import java.util.*;  public class GFG {      // Function that prints    // the Mode values    public static void findMode(int[] a, int n)    {        // Map used to map integers        // to its frequency        Map map            = new HashMap<>();          // To store the maximum frequency        int max = 0;          // To store the element with        // the maximum frequency        int mode = 0;          // Loop used to read the        // elements one by one        for (int i = 0; i < n; i++) {              // Updates the frequency of            // that element            map.put(a[i],                    map.getOrDefault(a[i], 0) + 1);              // Checks for maximum Number            // of occurrence            if (map.get(a[i]) >= max) {                  // Updates the maximum frequency                max = map.get(a[i]);                  // Updates the Mode                mode = a[i];            }              System.out.print(mode);            System.out.print(" ");        }    }      // Driver Code    public static void main(String[] args)    {        int arr[] = { 2, 7, 3, 2, 5 };          int n = arr.length;          // Function Call        findMode(arr, n);    }}

Python3

 # Python3 implementation of the # above approach  # Function that prints # the Mode values def findMode(a, n):          # Map used to mp integers     # to its frequency     mp = {}          # To store the maximum frequency     max = 0          # To store the element with     # the maximum frequency     mode = 0          # Loop used to read the     # elements one by one     for i in range(n):        if a[i] in mp:            mp[a[i]] += 1        else:            mp[a[i]] = 1                  # Checks for maximum Number         # of occurrence             if (mp[a[i]] >= max):                          # Updates the maximum             # frequency             max = mp[a[i]]                          # Updates the Mode             mode = a[i]                      print(mode, end = " ")  # Driver Codearr = [ 2, 7, 3, 2, 5 ]n = len(arr)  # Function call findMode(arr,n)  # This code is contributed by divyeshrabadiya07

C#

 // C# implementation of the// above approachusing System;using System.Collections.Generic;class GFG{      // Function that prints    // the Mode values    public static void findMode(int[] a, int n)    {        // Map used to map integers        // to its frequency        Dictionary map = new Dictionary();          // To store the maximum frequency        int max = 0;          // To store the element with        // the maximum frequency        int mode = 0;          // Loop used to read the        // elements one by one        for (int i = 0; i < n; i++)         {                      // Updates the frequency of            // that element            if (map.ContainsKey(a[i]))             {                map[a[i]] = map[a[i]] + 1;            }            else            {                map.Add(a[i], 1);            }              // Checks for maximum Number            // of occurrence            if (map[a[i]] >= max)             {                  // Updates the maximum frequency                max = map[a[i]];                  // Updates the Mode                mode = a[i];            }            Console.Write(mode);            Console.Write(" ");        }    }      // Driver Code    public static void Main(String[] args)    {        int[] arr = {2, 7, 3, 2, 5};        int n = arr.Length;          // Function Call        findMode(arr, n);    }}  // This code is contributed by Amit Katiyar
Output:
2 7 3 2 2

Performance Analysis:

• Time Complexity: O(N)
• Auxiliary Space: O(N)

My Personal Notes arrow_drop_up