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.
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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.Â
- 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;Â
string removeX(string N, char X){       // Stores the index of X    // that has to be removed    int index = -1;Â
    // 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') {Â
            // Update index and break            index = i;            break;        }    }Â
    // 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;            }        }    }Â
    // 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];    }Â
    return ans;}Â
int main(){Â
    string N = "2342";    char X = '2';    cout << removeX(N, X) << endl;    return 0;}Â
// This code is contributed by Ishan Khandelwal |
Java
// Java code to implement the approachÂ
import java.io.*;Â
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;Â
        // 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;            }        }Â
        // 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;                }            }        }Â
        // 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);        }Â
        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;Â
    # 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')):Â
            # Update index and break            index = i;            break;Â
    # 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;             Â
    # Construct answer using all characters    # in string N except index    ans = "";    for i in range(len(N)):        if (i != index):            ans = ans + N[i];     Â
    return ans;Â
N = "2342";X = '2';print(removeX(N, X));Â
# This code is contributed by Saurabh Jaiswal |
C#
// C# program for the above approachusing System;using System.Collections.Generic;Â
class GFG{Â
  // 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;Â
    // 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') {Â
        // Update index and break        index = i;        break;      }    }Â
    // 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;        }      }    }Â
    // 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];    }Â
    return ans;  } Â
  // Driver Code  public static void Main()  {    string N = "2342";    char X = '2';Â
    // Function call    Console.Write(removeX(N, X));  }}Â
// 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;Â
    // 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') {Â
            // Update index and break            index = i;            break;        }    }Â
    // 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;            }        }    }Â
    // 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];    }Â
    return ans;}Â
let N = "2342";let X = '2';document.write(removeX(N, X));Â
// This code is contributed by Samim Hossain Mondal.</script> |
Output
342
Time Complexity: O(N)
Auxiliary Space: O(1)
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