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Mrac [35]
2 years ago
8

Given an array of integers a, your task is to calculate the digits that occur the most number of times in the array. Return the

array of these digits in ascending order.
Computers and Technology
1 answer:
Annette [7]2 years ago
3 0

Answer:

I hope this should be helpful

Explanation:

In Javascript

function solution(numbers) {

   let hashmap = {};

   let ans = [];

   for (let i = 0; i < numbers.length; i++) {

     // check wether it double or single digit

     if (hasOneDigit(numbers[i])) {

       // check if it exists in hashmap

       // if exist add number of occurence

       // check if the occurence is greater than 2

       // add  if > 2 to the answer list

       if (numbers[i] in hashmap) {

         hashmap[numbers[i]] = hashmap[numbers[i]] + 1;

 

         if (hashmap[numbers[i]] = 2 && !ans.includes(numbers[i])) {

           // max_freq = hashmap[numbers[i]]

           ans.push(numbers[i]);

         }

         // else if (max_freq == hashmap[numbers[i]])

         //     ans = min(ans, numbers[i])

       } else {

         hashmap[numbers[i]] = 1;

       }

     } else {

       // change number to string

       number = numbers[i].toString();

       // loop to iterate every single element

       for (let j = 0; j < number.length; j += 1) {

 

         if (+number.charAt(j) in hashmap) {

           hashmap[+number.charAt(j)] = hashmap[+number.charAt(j)] + 1;

 

           if (hashmap[+number.charAt(j)] = 2 && !ans.includes(+number.charAt(j))) {

     

             ans.push(+number.charAt(j));

           }

       

         } else {

             hashmap[+number.charAt(j)] = 1

         }

       }

     }

   }

     return ans.sort();

 }

 

 function hasOneDigit(val) {

   return String(Math.abs(val)).charAt(0) == val;

 }

 console.log(solution([2, 1,42,44, 2, 3,32,7777, 3 , -1]));

 

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In Java, the following will work.
I made it a bit more versatile to work with others numbers, other than 99, if you so please (if not, just hardcode the 99 in yourself).

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Instructions The population of town A is less than the population of town B. However, the population of town A is growing faster
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Answer: The c++ program is given below.

#include <iostream>

using namespace std;

int main() {

float townA, townB, growthA, growthB, populationA, populationB;

    int year=0;    

cout<<"Enter present population of town A : ";

cin >> townA;  

cout<<endl<<"Enter present growth rate of town A : ";

cin >> growthA;

growthA = growthA/100;  

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cin >> townB;  

cout<<endl<<"Enter present growth rate of town B : ";

cin >> growthB;

growthB = growthB/100;  

do

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return 0;

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The growth rate entered by the user is the percentage growth.

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The program converts 10% into float as 10/100.

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The growth in population is computed in a do-while loop. After each growth is calculated, the year variable is incremented by 1.

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Once the loop discontinues, the final populations of town A and town B and the years needed for this growth is displayed.

The new line is introduced using endl keyword.

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