Generate string after adding spaces at specific positions in a given String

Given a string s and an array spaces[] describing the original string’s indices where spaces will be added. The task is to add spaces in given positions in spaces[] and print the string formed.

Examples:

Input: s = “GeeksForGeeK”, spaces = {1, 5, 10}
Output: “G eeks ForGe eK”    
Explanation: The underlined characters in “GeeksForGeeK” relate to the indices 1, 5, and 10. After that, put spaces in front of those characters.

Input: s = “ilovegeeksforgeek”, spaces = {1, 5, 10, 13}
Output: “i love geeks for geek”

Approach#1: This problem is simple string implementation based. Follow the steps below to solve the given problem. 

• Initialize with a space in the new string of size of the sum of the length of both arrays.
• Visit the index and wherever the index is equal to the current space value in the space[] array skip it as space is already there.
• Else keep assigning the value from the main string
• Here addition of ‘l’ in line { if(l<N and i==sp[l]+l) } is very interesting to observe:
  • The values in the space array are basically according to the old input string.
  • But in the new string, these space values or indices are basically shifting by the number of spaces found before.
• Print the string formed at the end.

Below is the implementation of the above approach

C++

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;

// Function to add space in required positions
string spaceintegration(string s, vector<int>& sp)
{
    int M = s.size(), N = sp.size(), l = 0, r = 0;

    string res(M + N, ' ');

    // Iterate over M+N length
    for (int i = 0; i < M + N; i++) {

        if (l < N and i == sp[l] + l)
            l++;
        else
            res[i] = s[r++];
    }

    // Return the required string
    return res;
}

// Driver Code
int main()
{

    string s = "ilovegeeksforgeeks";

    vector<int> space = { 1, 5, 10, 13 };

    // Function Call
    cout << spaceintegration(s, space) << endl;

    return 0;
}

Java

// Java program for above approach
import java.util.*;
class GFG
{

  // Function to add space in required positions
  static String spaceintegration(String s, int []sp)
  {
    int M = s.length(), N = sp.length, l = 0, r = 0;
    String res = newstr(M + N, ' ');

    // Iterate over M+N length
    for (int i = 0; i < M + N; i++) {

      if (l < N && i == sp[l] + l)
        l++;
      else
        res = res.substring(0,i)+s.charAt(r++)+res.substring(i+1);
    }

    // Return the required String
    return res;
  }

  static String newstr(int i, char c) {
    String str = "";
    for (int j = 0; j < i; j++) {
      str+=c;
    }
    return str;
  }

  // Driver Code
  public static void main(String[] args)
  {
    String s = "ilovegeeksforgeeks";
    int[] space = { 1, 5, 10, 13 };

    // Function Call
    System.out.print(spaceintegration(s, space) +"\n");

  }
}

// This code contributed by shikhasingrajput

Python3

# Python3 program for above approach

# Function to add space in required positions
def spaceintegration(s, sp):

    M = len(s)
    N = len(sp)
    l = 0
    r = 0

    res = [' '] * (M + N)

    # Iterate over M+N length
    for i in range(M + N):
        if (l < N and i == sp[l] + l):
            l += 1
        else:
            res[i] = s[r]
            r += 1

    # Return the required string
    return ''.join(res)

# Driver Code
if __name__ == "__main__":

    s = "ilovegeeksforgeeks"

    space = [ 1, 5, 10, 13 ]

    # Function Call
    print(spaceintegration(s, space))

# This code is contributed by ukasp

C#

// C# program for above approach
using System;
class GFG
{

  // Function to add space in required positions
  static String spaceintegration(String s, int []sp)
  {
    int M = s.Length, N = sp.Length, l = 0, r = 0;
    String res = newstr(M + N, ' ');

    // Iterate over M+N length
    for (int i = 0; i < M + N; i++) {

      if (l < N && i == sp[l] + l)
        l++;
      else
        res = res.Substring(0,i)+s[r++]+res.Substring(i+1);
    }

    // Return the required String
    return res;
  }

  static String newstr(int i, char c) {
    String str = "";
    for (int j = 0; j < i; j++) {
      str+=c;
    }
    return str;
  }

  // Driver Code
  public static void Main()
  {
    String s = "ilovegeeksforgeeks";
    int[] space = { 1, 5, 10, 13 };

    // Function Call
    Console.Write(spaceintegration(s, space) +"\n");

  }
}

// This code is contributed by Saurabh Jaiswal

Javascript

<script>
      // JavaScript code for the above approach

      // Function to add space in required positions
      function spaceintegration(s, sp)
      {
          let M = s.length, N = sp.length, l = 0, r = 0;
          let res = new Array(M + N).fill(' ');

          // Iterate over M+N length
          for (let i = 0; i < M + N; i++) {

              if (l < N && i == sp[l] + l)
                  l++;
              else
                  res[i] = s[r++];
          }

          // Return the required string
          return res.join('');
      }

      // Driver Code
      let s = "ilovegeeksforgeeks";
      let space = [1, 5, 10, 13];

      // Function Call
      document.write(spaceintegration(s, space) + '<br>');

// This code is contributed by Potta Lokesh
  </script>

Output

i love geeks for geeks

Time Complexity: O(M+N) 
Auxiliary Space: O(M+N)

Approach#2: This problem can be solve by method which we use to insert element at specific position in array. Follow the steps below to solve the given problem. 

• We have string s and position in vector space. 
• Iterate over the element of the space From last of vector to first and follow the following steps for every element of vector. Let L element of the vector.
  • Add one space at the end of s.
  • Iterate over string Till the L: 
    • Move one-one character forward till L.
  • At last Add space at L.
• Print string at the end.

Below is the implementation of the above approach.

C++

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;

// Function to add space in required positions
string spaceintegration(string s, vector<int>& space)
{
    int y = 0;
    int len = space.size();

    while (len--) {
        int k = space[len] + 1;
        int l = s.size() - 1;
        string tem = " ";
        s += tem;
        //   iterate over string
        for (int i = l; i >= k - 1; i--) {
            s[i + 1] = s[i];
        }
        s[k - 1] = tem[0];
        y += 1;
    }
    return s;
}

// Driver code
int main()
{

    string s = "ilovegeeksforgeeks";
    vector<int> space = { 1, 5, 10, 13 };

    //   Function call
    cout << spaceintegration(s, space) << endl;

    return 0;
}

Java

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

class GFG {
    // Function to add space in required positions
    public static String spaceIntegration(String s, List<Integer> space) {
        int len = space.size();
        while (len-- > 0) {
            int k = space.get(len) + 1;
            String tem = " ";
            s = s.substring(0, k - 1) + tem + s.substring(k - 1);
        }
        return s;
    }

    // Driver code
    public static void main(String[] args) {
        String s = "GeeksForGeeK";
        List<Integer> space = new ArrayList<Integer>();
        Collections.addAll(space, 1, 5, 10);

        // Function call
        System.out.println(spaceIntegration(s, space));
    }
}

Python

# Python3 program for above approach

# Function to add space in required positions


def spaceintegration(se, space):
    s = list(se)
    # Iterate over the string
    for i in range(len(space)-1, -1, -1):
        s.insert(space[i], " ")

    return "".join(s)


# Driver Code
if __name__ == "__main__":

    s = "ilovegeeksforgeeks"

    space = [1, 5, 10, 13]

    # Function Call
    print(spaceintegration(s, space))

# This code is contributed by ukasp

C#

// C# program to implement above approach
using System;
using System.Collections;
using System.Collections.Generic;

class GFG
{

    // Function to add space in required positions
    static string spaceintegration(string s, List<int> space)
    {
        int y = 0;
        int len = space.Count;

        while (len > 0) {
            len -= 1;
            int k = space[len] + 1;
            int l = s.Length - 1;
            string tem = " ";
            s += tem;

            // iterate over string
            for (int i = l - 1 ; i >= k - 1 ; i--) {
                  // Replaces s[i + 1] with s[i]
                s = s.Remove(i + 1, 1);
                s = s.Insert(i + 1, Char.ToString(s[i]));
            }
              // Replaces s[k - 1] with tem[0]
            s = s.Remove(k - 1, 1);
            s = s.Insert(k - 1, Char.ToString(tem[0]));
            y += 1;
        }
        return s;
    }

    // Driver code
    public static void Main(string[] args){

        string s = "ilovegeeksforgeeks";
        List<int> space = new List<int>{ 1, 5, 10, 13 };

        // Function call
        Console.WriteLine(spaceintegration(s, space));

    }
}

// This code is contributed by subhamgoyal2014.

Javascript

// JavaScript code for the above approach

        // Function to add space in required positions
        function spaceintegration(s, sp)
        {
           s = s.split('')
        //   iterate over the space
           for(let i = sp.length-1; i>=0; i--){
               s.splice(sp[i], 0, " ");
           }

            // Return the required string
            return s.join('');
        }

        // Driver Code
        let s = "ilovegeeksforgeeks";
        let space = [1, 5, 10, 13];

        // Function Call
        console.log(spaceintegration(s,space));

  // This code is contributed by Sam snehil

Output

i love geeks for geeks

Time Complexity: O(M*N) Here M is the length of the string and N is the size of space vector.
Auxiliary Space: O(1), As constant extra space is used.