Find the largest positive integer that can be formed by deleting only one occurrence of a given digit.
Examples:
Input: num = 56321, digit = 5
Output: 6321
Explanation: Since the number 56321 contain only 1 occurrence of 5, we can remove it to get 6321 which is the largest possible positive number.Input: num = 936230, digit = 3
Output: 96230
Explanation: Since the number 936230 contain 2 occurrences of 3, we can remove either 1st occurrence to get 96230 or remove 2nd occurrence to get 93620. Among the both, 96230 is the largest possible positive number.
Approach: The problem can be solved based on the following idea:
To find the maximum number, delete an occurrence in such a position where the next digit is greater than it. Because then the number formed will be large. Such a position should be as close to left as possible for higher values.
Follow the steps mentioned below to implement the idea:
- Remove the leftmost occurrence of X if it’s followed by a larger digit.
- If no occurrence of X is followed by a digit greater than X then remove the last occurrence of the digit.
Below is the implementation of the above approach.
C++
#include <bits/stdc++.h>using namespace std;string removeX(string N, char X){ // Stores the index of X // that has to be removed int index = -1; // Find leftmost occurrence of X // such that the digit just after X // is greater than X for (int i = 0; i < N.length() - 1; i++) { if (N[i] == X && N[i] - '0' < N[i + 1] - '0') { // Update index and break index = i; break; } } // If no occurrence of X such that // the digit just after X // is greater than X is found // then find last occurrence of X if (index == -1) { for (int i = N.length() - 1; i >= 0; i--) { if (N[i] == X) { index = i; break; } } } // Construct answer using all characters // in string N except index string ans = ""; for (int i = 0; i < N.length(); i++) { if (i != index) ans = ans + N[i]; } return ans;}int main(){ string N = "2342"; char X = '2'; cout << removeX(N, X) << endl; return 0;}// This code is contributed by Ishan Khandelwal |
Java
// Java code to implement the approachimport java.io.*;class GFG { // Function to find the largest number public static String removeX(String N, char X) { // Stores the index of X // that has to be removed int index = -1; // Find leftmost occurrence of X // such that the digit just after X // is greater than X for (int i = 0; i < N.length() - 1; i++) { if (N.charAt(i) == X && N.charAt(i) - '0' < N.charAt(i + 1) - '0') { // Update index and break index = i; break; } } // If no occurrence of X such that // the digit just after X // is greater than X is found // then find last occurrence of X if (index == -1) { for (int i = N.length() - 1; i >= 0; i--) { if (N.charAt(i) == X) { index = i; break; } } } // Construct answer using all characters // in string N except index String ans = ""; for (int i = 0; i < N.length(); i++) { if (i != index) ans = ans + N.charAt(i); } return ans; } // Driver code public static void main(String[] args) { String N = "2342"; char X = '2'; // Function call System.out.println(removeX(N, X)); }} |
Python3
# Python code to implement the approachdef removeX(N, X): # Stores the index of X # that has to be removed index = -1; # Find leftmost occurrence of X # such that the digit just after X # is greater than X for i in range(len(N) - 1): if (N[i] == X and ord(N[i]) - ord('0') < ord(N[i + 1]) - ord('0')): # Update index and break index = i; break; # If no occurrence of X such that # the digit just after X # is greater than X is found # then find last occurrence of X if (index == -1): for i in range(len(N), -1, -1): if (N[i] == X): index = i; break; # Construct answer using all characters # in string N except index ans = ""; for i in range(len(N)): if (i != index): ans = ans + N[i]; return ans;N = "2342";X = '2';print(removeX(N, X));# This code is contributed by Saurabh Jaiswal |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{ // Function to find the largest number public static string removeX(string N, char X) { // Stores the index of X // that has to be removed int index = -1; // Find leftmost occurrence of X // such that the digit just after X // is greater than X for (int i = 0; i < N.Length - 1; i++) { if (N[i] == X && N[i] - '0' < N[i + 1] - '0') { // Update index and break index = i; break; } } // If no occurrence of X such that // the digit just after X // is greater than X is found // then find last occurrence of X if (index == -1) { for (int i = N.Length - 1; i >= 0; i--) { if (N[i] == X) { index = i; break; } } } // Construct answer using all characters // in string N except index string ans = ""; for (int i = 0; i < N.Length; i++) { if (i != index) ans = ans + N[i]; } return ans; } // Driver Code public static void Main() { string N = "2342"; char X = '2'; // Function call Console.Write(removeX(N, X)); }}// This code is contributed by sanjoy_62. |
Javascript
<script>// Javascript code to implement the approachfunction removeX(N, X){ // Stores the index of X // that has to be removed let index = -1; // Find leftmost occurrence of X // such that the digit just after X // is greater than X for (let i = 0; i < N.length - 1; i++) { if (N[i] == X && N[i] - '0' < N[i + 1] - '0') { // Update index and break index = i; break; } } // If no occurrence of X such that // the digit just after X // is greater than X is found // then find last occurrence of X if (index == -1) { for (let i = N.length - 1; i >= 0; i--) { if (N[i] == X) { index = i; break; } } } // Construct answer using all characters // in string N except index let ans = ""; for (let i = 0; i < N.length; i++) { if (i != index) ans = ans + N[i]; } return ans;}let N = "2342";let X = '2';document.write(removeX(N, X));// This code is contributed by Samim Hossain Mondal.</script> |
Output
342
Time Complexity: O(N)
Auxiliary Space: O(1)
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