Minimum number to be added to all digits of X to make X > Y Last Updated : 08 Mar, 2022 Comments Improve Suggest changes Like Article Like Report Given two numbers X and Y of the same length, the task is to find the minimum number d that needs to be added to all the digits of X such that it becomes larger than Y.Examples: Input: X = 123, Y = 222 Output: 1 Explanation: Add 1 to all digits of X Then X becomes {2, 3, 4} which is lexicographically larger than {2, 2, 2}.Input: X = 4512, Y = 2998 Output: 0 Explanation: X is already lexicographically larger than Y so the answer will be 0. Approach: This problem can be solved easily by breaking it into three cases Case 1: Find if X is already lexicographically larger than Y. If yes, then we don't need to do anything.Case 2: Otherwise add (Y[0] - X[0]) to all elements of X and then check if X is lexicographically larger than Y or not.Case 3: If it is still not larger, then answer will be (Y[0] - X[0]) + 1 because the first elements of X become larger than first element of Y means X[0] > Y[0]. Below is the implementation of the above approach: C++ // C++ program to find Minimum number to be added // to all digits of X to make X > Y #include <bits/stdc++.h> using namespace std; // Function to check if X // is lexicographically larger Y bool IsLarger(int X[], int Y[], int n) { for (int i = 0; i < n; i++) { // It is lexicographically larger if (X[i] < Y[i]) { return false; } } return true; } // Utility function to check // minimum value of d int solve(int X[], int Y[], int n) { int ans = 0; // If X is already larger // do not need to add anything if (IsLarger(X, Y, n)) { ans = 0; } else { // Adding d to all elements of X int d = Y[0] - X[0]; for (int i = 0; i < n; i++) { X[i] += d; } // If X is larger now // print d if (IsLarger(X, Y, n)) { ans = d; } // else print d + 1 else { ans = d + 1; } } return ans; } // Driver Code int main() { // Taking the numbers as sequences int X[] = { 2, 3, 6, 9 }; int Y[] = { 3, 4, 8, 1 }; int n = sizeof(X) / sizeof(X[0]); cout << solve(X, Y, n); return 0; } Java // Java program to find Minimum number to be added // to all digits of X to make X > Y import java.util.*; class GFG { // Function to check if X // is lexicographically larger Y static boolean IsLarger(int[] X, int[] Y, int n) { for (int i = 0; i < n; i++) { // It is lexicographically larger if (X[i] < Y[i]) { return false; } } return true; } // Utility function to check // minimum value of d static int solve(int X[], int Y[], int n) { int ans = 0; // If X is already larger // do not need to add anything if (IsLarger(X, Y, n)) ans = 0; else { // Adding d to all elements of X int d = Y[0] - X[0]; for (int i = 0; i < n; i++) X[i] += d; // If X is larger now // print d if (IsLarger(X, Y, n)) ans = d; // else print d + 1 else { ans = d + 1; } } return ans; } // Driver Code public static void main(String[] args) { // Taking the numbers as sequences int X[] = { 2, 3, 6, 9 }; int Y[] = { 3, 4, 8, 1 }; int n = X.length; System.out.println(solve(X, Y, n)); } } // This code is contributed by // sanjeev2552 Python3 # Python3 program to find Minimum number to be added # to all digits of X to make X > Y # Function to check if X # is lexicographically larger Y def IsLarger(X, Y, n) : for i in range(n) : # It is lexicographically larger if (X[i] < Y[i]) : return False; return True; # Utility function to check # minimum value of d def solve(X, Y, n) : ans = 0; # If X is already larger # do not need to add anything if (IsLarger(X, Y, n)) : ans = 0; else : # Adding d to all elements of X d = Y[0] - X[0]; for i in range(n) : X[i] += d; # If X is larger now # print d if (IsLarger(X, Y, n)) : ans = d; # else print d + 1 else : ans = d + 1; return ans; # Driver Code if __name__ == "__main__" : # Taking the numbers as sequences X = [ 2, 3, 6, 9 ]; Y = [ 3, 4, 8, 1 ]; n = len(X); print(solve(X, Y, n)); # This code is contributed by AnkitRai01 C# // C# program to find Minimum number to be.Added // to all digits of X to make X > Y using System; class GFG { // Function to check if X // is lexicographically larger Y static bool IsLarger(int[] X, int[] Y, int n) { for (int i = 0; i < n; i++) { // It is lexicographically larger if (X[i] < Y[i]) { return false; } } return true; } // Utility function to check // minimum value of d static int solve(int []X, int []Y, int n) { int ans = 0; // If X is already larger // do not need to.Add anything if (IsLarger(X, Y, n)) ans = 0; else { // Adding d to all elements of X int d = Y[0] - X[0]; for (int i = 0; i < n; i++) X[i] += d; // If X is larger now // print d if (IsLarger(X, Y, n)) ans = d; // else print d + 1 else { ans = d + 1; } } return ans; } // Driver Code public static void Main(String[] args) { // Taking the numbers as sequences int []X = { 2, 3, 6, 9 }; int []Y = { 3, 4, 8, 1 }; int n = X.Length; Console.WriteLine(solve(X, Y, n)); } } // This code is contributed by PrinciRaj1992 JavaScript <script> // javascript program to find Minimum number to be added // to all digits of X to make X > Y // Function to check if X // is lexicographically larger Y function IsLarger(X, Y, n) { for (let i = 0; i < n; i++) { // It is lexicographically larger if (X[i] < Y[i]) { return false; } } return true; } // Utility function to check // minimum value of d function solve(X, Y, n) { let ans = 0; // If X is already larger // do not need to add anything if (IsLarger(X, Y, n)) { ans = 0; } else { // Adding d to all elements of X let d = Y[0] - X[0]; for (let i = 0; i < n; i++) { X[i] += d; } // If X is larger now // print d if (IsLarger(X, Y, n)) { ans = d; } // else print d + 1 else { ans = d + 1; } } return ans; } // Driver Code // Taking the numbers as sequences let X = [ 2, 3, 6, 9 ]; let Y = [ 3, 4, 8, 1 ]; let n = X.length; document.write(solve(X, Y, n)); // This code is contributed by souravmahato348. </script> Output: 2 Time Complexity: O(N) , where N is the length of X or Y Auxiliary Space: O(1) Comment More infoAdvertise with us Next Article Analysis of Algorithms S souradeep Follow Improve Article Tags : Greedy Mathematical DSA Constructive Algorithms Practice Tags : GreedyMathematical Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. 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