Find the Largest number with given number of digits and sum of digits
How to find the largest number with given digit sum s and number of digits d?
Examples:
Input : s = 9, d = 2 Output : 90 Input : s = 20, d = 3 Output : 992
A Simple Solution is to consider all m digit numbers and keep track of maximum number with digit sum as s. A close upper bound on time complexity of this solution is O(10m).
There is a Greedy approach to solve the problem. The idea is to one by one fill all digits from leftmost to rightmost (or from most significant digit to least significant).
We compare the remaining sum with 9 if the remaining sum is more than 9, we put 9 at the current position, else we put the remaining sum. Since we fill digits from left to right, we put the highest digits on the left side, hence get the largest number.
Below image is an illustration of the above approach:
Below is the implementation of the above approach:
C++
// C++ program to find the largest number that can be// formed from given sum of digits and number of digits.#include <iostream>using namespace std;// Prints the smalles possible number with digit sum 's'// and 'm' number of digits.void findLargest(int m, int s){ // If sum of digits is 0, then a number is possible // only if number of digits is 1. if (s == 0) { (m == 1)? cout << "Largest number is " << 0 : cout << "Not possible"; return ; } // Sum greater than the maximum possible sum. if (s > 9*m) { cout << "Not possible"; return ; } // Create an array to store digits of result int res[m]; // Fill from most significant digit to least // significant digit. for (int i=0; i<m; i++) { // Fill 9 first to make the number largest if (s >= 9) { res[i] = 9; s -= 9; } // If remaining sum becomes less than 9, then // fill the remaining sum else { res[i] = s; s = 0; } } cout << "Largest number is "; for (int i=0; i<m; i++) cout << res[i];}// Driver codeint main(){ int s = 9, m = 2; findLargest(m, s); return 0;} |
Java
// Java program to find the largest number that can be// formed from given sum of digits and number of digitsclass GFG{ // Function to print the largest possible number with digit sum 's' // and 'm' number of digits static void findLargest(int m, int s) { // If sum of digits is 0, then a number is possible // only if number of digits is 1 if (s == 0) { System.out.print(m == 1 ? "Largest number is 0" : "Not possible"); return ; } // Sum greater than the maximum possible sum if (s > 9*m) { System.out.println("Not possible"); return ; } // Create an array to store digits of result int[] res = new int[m]; // Fill from most significant digit to least // significant digit for (int i=0; i<m; i++) { // Fill 9 first to make the number largest if (s >= 9) { res[i] = 9; s -= 9; } // If remaining sum becomes less than 9, then // fill the remaining sum else { res[i] = s; s = 0; } } System.out.print("Largest number is "); for (int i=0; i<m; i++) System.out.print(res[i]); } // driver program public static void main (String[] args) { int s = 9, m = 2; findLargest(m, s); }}// Contributed by Pramod Kumar |
Python3
# Python 3 program to find# the largest number that# can be formed from given# sum of digits and number# of digits.# Prints the smalles# possible number with digit# sum 's' and 'm' number of# digits.def findLargest( m, s) : # If sum of digits is 0, # then a number is possible # only if number of digits # is 1. if (s == 0) : if(m == 1) : print("Largest number is " , "0",end = "") else : print("Not possible",end = "") return # Sum greater than the # maximum possible sum. if (s > 9 * m) : print("Not possible",end = "") return # Create an array to # store digits of # result res = [0] * m # Fill from most significant # digit to least significant # digit. for i in range(0, m) : # Fill 9 first to make # the number largest if (s >= 9) : res[i] = 9 s = s - 9 # If remaining sum # becomes less than # 9, then fill the # remaining sum else : res[i] = s s = 0 print( "Largest number is ",end = "") for i in range(0, m) : print(res[i],end = "")# Driver codes = 9m = 2findLargest(m, s)# This code is contributed by Nikita Tiwari. |
C#
// C# program to find the// largest number that can// be formed from given sum// of digits and number of digitsusing System;class GFG{ // Function to print the // largest possible number // with digit sum 's' and // 'm' number of digits static void findLargest(int m, int s) { // If sum of digits is 0, // then a number is possible // only if number of digits is 1 if (s == 0) { Console.Write(m == 1 ? "Largest number is 0" : "Not possible"); return ; } // Sum greater than the // maximum possible sum if (s > 9 * m) { Console.WriteLine("Not possible"); return ; } // Create an array to // store digits of result int []res = new int[m]; // Fill from most significant // digit to least significant digit for (int i = 0; i < m; i++) { // Fill 9 first to make // the number largest if (s >= 9) { res[i] = 9; s -= 9; } // If remaining sum becomes // less than 9, then // fill the remaining sum else { res[i] = s; s = 0; } } Console.Write("Largest number is "); for (int i = 0; i < m; i++) Console.Write(res[i]); } // Driver Code static public void Main () { int s = 9, m = 2; findLargest(m, s); }}// This code is Contributed by ajit |
PHP
<?php// PHP program to find the largest// number that can be formed from// given sum of digits and number// of digits.// Prints the smalles possible// number with digit sum 's'// and 'm' number of digits.function findLargest($m, $s){ // If sum of digits is 0, then // a number is possible only if // number of digits is 1. if ($s == 0) { if(($m == 1) == true) echo "Largest number is " , 0; else echo "Not possible"; return ; } // Sum greater than the // maximum possible sum. if ($s > 9 * $m) { echo "Not possible"; return ; } // Create an array to store // digits of result Fill from // most significant digit to // least significant digit. for ($i = 0; $i < $m; $i++) { // Fill 9 first to make // the number largest if ($s >= 9) { $res[$i] = 9; $s -= 9; } // If remaining sum becomes // less than 9, then fill // the remaining sum else { $res[$i] = $s; $s = 0; } } echo "Largest number is "; for ($i = 0; $i < $m; $i++) echo $res[$i];}// Driver code$s = 9; $m = 2;findLargest($m, $s);// This code is contributed by m_kit?> |
Javascript
<script>// Javascript program to find the largest number that can be// formed from given sum of digits and number of digits.// Prints the smalles possible number with digit sum 's'// and 'm' number of digits.function findLargest(m, s){ // If sum of digits is 0, then a number is possible // only if number of digits is 1. if (s == 0) { (m == 1)? document.write("Largest number is " + 0) : document.write("Not possible"); return ; } // Sum greater than the maximum possible sum. if (s > 9*m) { document.write("Not possible"); return ; } // Create an array to store digits of result let res = new Array(m); // Fill from most significant digit to least // significant digit. for (let i=0; i<m; i++) { // Fill 9 first to make the number largest if (s >= 9) { res[i] = 9; s -= 9; } // If remaining sum becomes less than 9, then // fill the remaining sum else { res[i] = s; s = 0; } } document.write("Largest number is "); for (let i=0; i<m; i++) document.write(res[i]);}// Driver code let s = 9, m = 2; findLargest(m, s);// This code is contributed by Mayank Tyagi</script> |
Output :
Largest number is 90
Time Complexity of this solution is O(m).
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