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# Minimize increments to make digit sum of N at most S

Given two positive integers N and S. The task is to minimize the number of increments on N (N = N + 1) required to make the sum of digits of N less than or equal to S
NOTE: N can range up to an18 digit number and 1 <= S <= 162.

Examples:

Input: N = 600, S = 5
Output: 400
Explanation: Minimum 400 increments are required as the sum of digits of 1000(600 + 400) is less than 5.

Input: N = 345899211156769, S = 20
Output: 100788843231
Explanation: Minimum required increments are 100788843231.

Approach: The key observation here is that in order to minimize the sum of digits of any number in minimum increments, the digits starting from the unit’s place need to be minimized. Follow the steps below to solve the given problem.

• For the base case, If the sum of digits of N is already less than or equal to S, then output is 0.
• Initialize variables, say ans =0 and p = 1, where ans will store the minimum required increments and p will keep track of the digits place, starting from the unit’s place.
• Run a loop through the maximum number of digits N can have (i.e., 18) and find the last digit of N. Let it be denoted by digit. Then find the minimum increments required to convert this digit to 0. Denoted by say, add.
• digit = (N / p) % 10
• add = p * (10 – digit)
• Increment N and ans by add.
• Now check whether the sum of digits of N is less than or equal to S.
• If the condition is true, break from the loop and output the answer.
• Otherwise, multiply p by 10 to access the second last digit from the unit’s place of N.
• The loop runs again and the same steps are executed till the required answer is found.
• Return ans as the final answer.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find the sum``// of digits of N``int` `findSum(``long` `long` `int` `N)``{``    ``// Stores the sum of digits of N``    ``int` `res = 0;` `    ``// Loop to extract the digits of N``    ``// and find their sum``    ``while` `(N) {` `        ``// Extracting the last digit of N``        ``// and adding it to res``        ``res += (N % 10);` `        ``// Update N``        ``N /= 10;``    ``}` `    ``return` `res;``}` `// Function to find the minimum increments``// required to make the sum of digits of N``// less than or equal to S.``long` `long` `int` `minIncrements(``long` `long` `int` `N, ``int` `S)``{``    ``// If the sum of digits of N is less than``    ``// or equal to S``    ``if` `(findSum(N) <= S) {` `        ``// Output 0``        ``return` `0;``    ``}` `    ``// variable to access the digits of N``    ``long` `long` `int` `p = 1;` `    ``// Stores the required answer``    ``long` `long` `int` `ans = 0;` `    ``// Loop to access the digits of N``    ``for` `(``int` `i = 0; i <= 18; ++i) {` `        ``// Stores the digit of N starting``        ``// from unit's place``        ``int` `digit = (N / p) % 10;` `        ``// Stores the increment required``        ``// to make the digit 0``        ``long` `long` `int` `add = p * (10 - digit);` `        ``// Update N``        ``N += add;` `        ``// Update ans``        ``ans += add;` `        ``// If the sum of digits of N is less than``        ``// or equal to S``        ``if` `(findSum(N) <= S) {` `            ``// Break from the loop``            ``break``;``        ``}` `        ``// Update p to access the next digit``        ``p = p * 10;``    ``}` `    ``return` `ans;``}` `// Driver Code``int` `main()``{` `    ``// Given N and S``    ``long` `long` `int` `N = 345899211156769;``    ``int` `S = 20;` `    ``// Function call``    ``cout << minIncrements(N, S);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;``public` `class` `GFG``{``  ` `// Function to find the sum``// of digits of N``static` `int` `findSum(``long` `N)``{``  ` `    ``// Stores the sum of digits of N``    ``int` `res = ``0``;` `    ``// Loop to extract the digits of N``    ``// and find their sum``    ``while` `(N != ``0``) {` `        ``// Extracting the last digit of N``        ``// and adding it to res``        ``res += (N % ``10``);` `        ``// Update N``        ``N /= ``10``;``    ``}` `    ``return` `res;``}` `// Function to find the minimum increments``// required to make the sum of digits of N``// less than or equal to S.``static` `long` `minIncrements(``long` `N, ``int` `S)``{``  ` `    ``// If the sum of digits of N is less than``    ``// or equal to S``    ``if` `(findSum(N) <= S) {` `        ``// Output 0``        ``return` `0``;``    ``}` `    ``// variable to access the digits of N``    ``long` `p = ``1``;` `    ``// Stores the required answer``    ``long` `ans = ``0``;` `    ``// Loop to access the digits of N``    ``for` `(``int` `i = ``0``; i <= ``18``; ++i) {` `        ``// Stores the digit of N starting``        ``// from unit's place``        ``long` `digit = (N / p) % ``10``;` `        ``// Stores the increment required``        ``// to make the digit 0``        ``long` `add = p * (``10` `- digit);` `        ``// Update N``        ``N += add;` `        ``// Update ans``        ``ans += add;` `        ``// If the sum of digits of N is less than``        ``// or equal to S``        ``if` `(findSum(N) <= S) {` `            ``// Break from the loop``            ``break``;``        ``}` `        ``// Update p to access the next digit``        ``p = p * ``10``;``    ``}` `    ``return` `ans;``}` `// Driver Code``public` `static` `void` `main(String args[])``{``    ``// Given N and S``    ``long` `N = 345899211156769L;``    ``int` `S = ``20``;` `    ``// Function call``    ``System.out.println(minIncrements(N, S));``}``}` `// This code is contributed by Samim Hossain Mondal`

## Python3

 `# Python program for the above approach` `# Function to find the sum``# of digits of N``def` `findSum(N):``  ` `  ``# Stores the sum of digits of N``  ``res ``=` `0``;` `  ``# Loop to extract the digits of N``  ``# and find their sum``  ``while` `(N):` `    ``# Extracting the last digit of N``    ``# and adding it to res``    ``res ``+``=` `(N ``%` `10``);` `    ``# Update N``    ``N ``=` `N ``/``/` `10``;``  ` `  ``return` `res;` `# Function to find the minimum increments``# required to make the sum of digits of N``# less than or equal to S.``def` `minIncrements(N, S):``  ` `  ``# If the sum of digits of N is less than``  ``# or equal to S``  ``if` `(findSum(N) <``=` `S):` `    ``# Output 0``    ``return` `0``;``  ` `  ``# variable to access the digits of N``  ``p ``=` `1``;` `  ``# Stores the required answer``  ``ans ``=` `0``;` `  ``# Loop to access the digits of N``  ``for` `i ``in` `range``(``0``, ``18``):` `    ``# Stores the digit of N starting``    ``# from unit's place``    ``digit ``=` `(N ``/``/` `p) ``%` `10``;` `    ``# Stores the increment required``    ``# to make the digit 0``    ``add ``=` `p ``*` `(``10` `-` `digit);` `    ``# Update N``    ``N ``+``=` `add;` `    ``# Update ans``    ``ans ``+``=` `add;` `    ``# If the sum of digits of N is less than``    ``# or equal to S``    ``if` `(findSum(N) <``=` `S):` `      ``# Break from the loop``      ``break``;``    ` `    ``# Update p to access the next digit``    ``p ``=` `p ``*` `10``;``  ` `  ``return` `ans;` `# Driver Code` `# Given N and S``N ``=` `345899211156769``;``S ``=` `20``;` `# Function call``print``(minIncrements(N, S))` `# This code is contributed by saurabh_jaiswal.`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections;` `class` `GFG``{``  ` `// Function to find the sum``// of digits of N``static` `long` `findSum(``long` `N)``{``  ` `    ``// Stores the sum of digits of N``    ``long` `res = 0;` `    ``// Loop to extract the digits of N``    ``// and find their sum``    ``while` `(N != 0) {` `        ``// Extracting the last digit of N``        ``// and adding it to res``        ``res += (N % 10);` `        ``// Update N``        ``N /= 10;``    ``}` `    ``return` `res;``}` `// Function to find the minimum increments``// required to make the sum of digits of N``// less than or equal to S.``static` `long` `minIncrements(``long` `N, ``long` `S)``{``  ` `    ``// If the sum of digits of N is less than``    ``// or equal to S``    ``if` `(findSum(N) <= S) {` `        ``// Output 0``        ``return` `0;``    ``}` `    ``// variable to access the digits of N``    ``long` `p = 1;` `    ``// Stores the required answer``    ``long` `ans = 0;` `    ``// Loop to access the digits of N``    ``for` `(``int` `i = 0; i <= 18; ++i) {` `        ``// Stores the digit of N starting``        ``// from unit's place``        ``long` `digit = (N / p) % 10;` `        ``// Stores the increment required``        ``// to make the digit 0``        ``long` `add = p * (10 - digit);` `        ``// Update N``        ``N += add;` `        ``// Update ans``        ``ans += add;` `        ``// If the sum of digits of N is less than``        ``// or equal to S``        ``if` `(findSum(N) <= S) {` `            ``// Break from the loop``            ``break``;``        ``}` `        ``// Update p to access the next digit``        ``p = p * 10;``    ``}``    ` `    ``return` `ans;``}` `// Driver Code``public` `static` `void` `Main()``{``    ``// Given N and S``    ``long` `N = 345899211156769;``    ``long` `S = 20;` `    ``// Function call``    ``Console.Write(minIncrements(N, S));``}``}` `// This code is contributed by Samim Hossain Mondal`

## Javascript

 ``

Output

`100788843231`

Time Complexity: O(18*logN).

Auxiliary Space: O(1).