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Remove last occurrence of a word from a given sentence string

  • Difficulty Level : Medium
  • Last Updated : 06 Aug, 2021

Given two strings S and W of sizes N and M respectively, the task is to remove the last occurrence of W from S. If there is no occurrence of W in S, print S as it is.

Examples:

Input: S = “This is GeeksForGeeks”, W=”Geeks”
Output: This is GeeksFor
Explanation:
The last occurrence of “Geeks” in the string is substring over the range [16, 20].

Input: S=”Hello World”, W=”Hell”
Output: o World
Explanation:
The last occurrence of “Hell” in the string is substring over the range [0, 3].

Approach: The problem can be solved by iterating over every index i of string S and checking if there is a substring starting from the index i, which is equal to string W. Follow the steps below to solve the problem:



  • If N is smaller than M, print S, as there can be no occurrence of W in S.
  • Initialize a variable i as N-M to iterate over the string S.
  • Iterate until i is greater than 0 and perform the following steps:
    • Check whether the substring over the range [i, i+M-1] is equal to string W or not. If it is equal, then remove the substring over the range [i, i+M-1] from string S and then break.
    • Otherwise, continue.
  • Finally, after completing the above steps, print the string S as the answer.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to remove last occurrence
// of W from S
string removeLastOccurrence(string S, string W, int N,
                            int M)
{
 
    // If M is greater than N
    if (M > N)
        return S;
 
    // Iterate while i is greater than
    // or equal to 0
    for (int i = N - M; i >= 0; i--) {
        // Stores if occurrence of W has
        // been found or not
        int flag = 0;
 
        // Iterate over the range [0, M]
        for (int j = 0; j < M; j++) {
 
            // If S[j+1] is not equal to
            // W[j]
            if (S[j + i] != W[j]) {
 
                // Mark flag true and break
                flag = 1;
                break;
            }
        }
 
        // If occurrence has been found
        if (flag == 0) {
 
            // Delete the substring over the
            // range [i, i+M]
            for (int j = i; j < N - M; j++)
                S[j] = S[j + M];
 
            // Resize the string S
            S.resize(N - M);
            break;
        }
    }
 
    // Return S
    return S;
}
// Driver Code
int main()
{
    // Input
    string S = "This is GeeksForGeeks";
    string W = "Geeks";
    int N = S.length();
    int M = W.length();
 
    // Function call
    cout << removeLastOccurrence(S, W, N, M) << endl;
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to remove last occurrence
// of W from S
static String removeLastOccurrence(String S, String W,
                                   int N, int M)
{
     
    // If M is greater than N
    char[] ch = S.toCharArray();
    if (M > N)
        return S;
 
    // Iterate while i is greater than
    // or equal to 0
    for(int i = N - M; i >= 0; i--)
    {
         
        // Stores if occurrence of W has
        // been found or not
        int flag = 0;
 
        // Iterate over the range [0, M]
        for(int j = 0; j < M; j++)
        {
             
            // If S[j+1] is not equal to
            // W[j]
            if (ch[j + i] != W.charAt(j))
            {
                 
                // Mark flag true and break
                flag = 1;
                break;
            }
        }
 
        // If occurrence has been found
        if (flag == 0)
        {
             
            // Delete the substring over the
            // range [i, i+M]
            for(int j = i; j < N - M; j++)
                ch[j] = ch[j + M];
                 
            break;
        }
    }
     
    char[] chh = new char[N - M];
     
    // Resize the string S
    for(int i = 0; i < N - M; i++)
    {
        chh[i] = ch[i];
    }
     
    // Return S
    return String.valueOf(chh);
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Input
    String S = "This is GeeksForGeeks";
    String W = "Geeks";
    int N = S.length();
    int M = W.length();
 
    // Function call
    System.out.print(removeLastOccurrence(S, W, N, M));
}
}
 
// This code is contributed by sanjoy_62

Python3




# Python3 program for the above approach
 
# Function to remove last occurrence
# of W from S
def removeLastOccurrence(S, W, N, M):
    S = [i for i in S]
    W = [i for i in W]
 
    # If M is greater than N
    if (M > N):
        return S
 
    # Iterate while i is greater than
    # or equal to 0
    for i in range(N - M, -1, -1):
        # of W has
        # been found or not
        flag = 0
 
        # Iterate over the range [0, M]
        for j in range(M):
            # If S[j+1] is not equal to
            # W[j]
            if (S[j + i] != W[j]):
 
                # Mark flag true and break
                flag = 1
                break
 
        # If occurrence has been found
        if (flag == 0):
 
            # Delete the subover the
            # range [i, i+M]
            for j in range(i,N-M):
                S[j] = S[j + M]
 
            # Resize the S
            S = S[:N - M]
            break
 
    # Return S
    return "".join(S)
   
# Driver Code
if __name__ == '__main__':
    # Input
    S = "This is GeeksForGeeks"
    W = "Geeks"
    N = len(S)
    M = len(W)
 
    # Function call
    print (removeLastOccurrence(S, W, N, M))
 
# This code is contributed by mohit kumar 29.

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG
{
   
// Function to remove last occurrence
// of W from S
static string removeLastOccurrence(string S, string W, int N,
                            int M)
{
 
    // If M is greater than N
   char[] ch = S.ToCharArray();
    if (M > N)
        return S;
 
    // Iterate while i is greater than
    // or equal to 0
    for (int i = N - M; i >= 0; i--)
    {
       
        // Stores if occurrence of W has
        // been found or not
        int flag = 0;
 
        // Iterate over the range [0, M]
        for (int j = 0; j < M; j++) {
 
            // If S[j+1] is not equal to
            // W[j]
            if (ch[j + i] != W[j]) {
 
                // Mark flag true and break
                flag = 1;
                break;
            }
        }
 
        // If occurrence has been found
        if (flag == 0) {
 
            // Delete the substring over the
            // range [i, i+M]
            for (int j = i; j < N - M; j++)
                ch[j] = ch[j + M];
 
            // Resize the string S
            Array.Resize(ref ch,N - M);
            break;
        }
    }
     S = string.Concat(ch);
   
    // Return S
    return S;
}
 
// Driver Code
public static void Main()
{
    // Input
    string S = "This is GeeksForGeeks";
    string W = "Geeks";
    int N = S.Length;
    int M = W.Length;
 
    // Function call
    Console.Write(removeLastOccurrence(S, W, N, M));
}
}
 
// This code is contributed by bgangwar59.
Output
This is GeeksFor

Time Complexity: O(M*N)
Auxiliary Space: O(1)

 

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