Given a number H which represent the total number of holes. The task is to find the smallest number which has that many numbers of holes.
NOTE:
- 0, 4, 6, 9 has 1 holes each and 8 has 2 holes in it.
- The number should not contain leading zeros.
Examples:
Input: H = 1
Output: 0
Input: H = 5
Output: 488
Explanation:
Number which has 5 holes in it is 488. i.e (1 + 2 + 2)
Refer: Count the number of holes in an integer
Approach:
- First of all, Check whether the number of holes given is 0 or 1, if 0 then print 1 and if 1 then print 0.
- If number of holes given is more than 1 then divide the number of holes by 2 and store the remainder in ‘rem’ variable. and quotient in ‘quo’ variable.
- Now, if value of rem variable is equal to 1 then first print 4 once and then print 8 quo number of times.
- Else print 8 only quo number of times.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function that will find out// the numbervoid printNumber(int holes){ // If number of holes // equal 0 then return 1 if (holes == 0) cout << "1"; // If number of holes // equal 0 then return 0 else if (holes == 1) cout << "0"; // If number of holes // is more than 0 or 1. else { int rem = 0, quo = 0; rem = holes % 2; quo = holes / 2; // If number of holes is // odd if (rem == 1) cout << "4"; for (int i = 0; i < quo; i++) cout << "8"; }}// Driver codeint main(){ int holes = 3; // Calling Function printNumber(holes); return 0;} |
Java
// Java implementation of the above approachimport java.io.*;class GFG { // Function that will find out// the numberstatic void printNumber(int holes){ // If number of holes // equal 0 then return 1 if (holes == 0) System.out.print("1"); // If number of holes // equal 0 then return 0 else if (holes == 1) System.out.print("0"); // If number of holes // is more than 0 or 1. else { int rem = 0, quo = 0; rem = holes % 2; quo = holes / 2; // If number of holes is // odd if (rem == 1) System.out.print("4"); for (int i = 0; i < quo; i++) System.out.print("8"); }}// Driver codepublic static void main (String[] args) { int holes = 3; // Calling Function printNumber(holes);}}// This code is contributed by Sachin. |
Python3
# Python3 implementation of # the above approach# Function that will find out# the numberdef printNumber(holes): # If number of holes # equal 0 then return 1 if (holes == 0): print("1") # If number of holes # equal 0 then return 0 elif (holes == 1): print("0", end = "") # If number of holes # is more than 0 or 1. else: rem = 0 quo = 0 rem = holes % 2 quo = holes // 2 # If number of holes is # odd if (rem == 1): print("4", end = "") for i in range(quo): print("8", end = "")# Driver codeholes = 3# Calling FunctionprintNumber(holes)# This code is contributed by Mohit kumar |
C#
// C# implementation of the above approachusing System;class GFG{ // Function that will find out// the numberstatic void printNumber(int holes){ // If number of holes // equal 0 then return 1 if (holes == 0) Console.Write ("1"); // If number of holes // equal 0 then return 0 else if (holes == 1) Console.Write ("0"); // If number of holes // is more than 0 or 1. else { int rem = 0, quo = 0; rem = holes % 2; quo = holes / 2; // If number of holes is // odd if (rem == 1) Console.Write ("4"); for (int i = 0; i < quo; i++) Console.Write ("8"); }}// Driver codestatic public void Main (){ int holes = 3; // Calling Function printNumber(holes);}}// This code is contributed by jit_t |
Javascript
<script> // Javascript implementation of the above approach // Function that will find out // the number function printNumber(holes) { // If number of holes // equal 0 then return 1 if (holes == 0) document.write("1"); // If number of holes // equal 0 then return 0 else if (holes == 1) document.write("0"); // If number of holes // is more than 0 or 1. else { let rem = 0, quo = 0; rem = holes % 2; quo = parseInt(holes / 2, 10); // If number of holes is // odd if (rem == 1) document.write("4"); for (let i = 0; i < quo; i++) document.write("8"); } } let holes = 3; // Calling Function printNumber(holes); // This code is contributed by divyeshrabadiya07.</script> |
Output:
48
Time Complexity: O(n)
Auxiliary Space: O(1)
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