Open In App
Related Articles

Recursive solution to count substrings with same first and last characters

Improve
Improve
Improve
Like Article
Like
Save Article
Save
Report issue
Report

We are given a string S, we need to find count of all contiguous substrings starting and ending with same character.

Examples : 

Input  : S = "abcab"
Output : 7
There are 15 substrings of "abcab"
a, ab, abc, abca, abcab, b, bc, bca
bcab, c, ca, cab, a, ab, b
Out of the above substrings, there 
are 7 substrings : a, abca, b, bcab, 
c, a and b.

Input  : S = "aba"
Output : 4
The substrings are a, b, a and aba

We have discussed different solutions in below post.
Count substrings with same first and last characters

In this article, a simple recursive solution is discussed. 

Implementation:

C++

// c++ program to count substrings with same
// first and last characters
#include <iostream>
#include <string>
using namespace std;
 
/* Function to count substrings with same first and
  last characters*/
int countSubstrs(string str, int i, int j, int n)
{
    // base cases
    if (n == 1)
        return 1;
    if (n <= 0)
        return 0;
 
    int res =  countSubstrs(str, i + 1, j, n - 1) + 
               countSubstrs(str, i, j - 1, n - 1) -
               countSubstrs(str, i + 1, j - 1, n - 2);           
 
    if (str[i] == str[j])
        res++;
 
    return res;
}
 
// driver code
int main()
{
    string str = "abcab";
    int n = str.length();
    cout << countSubstrs(str, 0, n - 1, n);
}

                    

Java

// Java program to count substrings
// with same first and last characters
 
class GFG
{
    // Function to count substrings
    // with same first and
    // last characters
    static int countSubstrs(String str, int i,
                                         int j, int n)
    {
        // base cases
        if (n == 1)
            return 1;
        if (n <= 0)
            return 0;
     
        int res = countSubstrs(str, i + 1, j, n - 1) +
                countSubstrs(str, i, j - 1, n - 1) -
                countSubstrs(str, i + 1, j - 1, n - 2);        
     
        if (str.charAt(i) == str.charAt(j))
            res++;
     
        return res;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        String str = "abcab";
        int n = str.length();
        System.out.print(countSubstrs(str, 0, n - 1, n));
    }
}
 
// This code is contributed by Anant Agarwal.

                    

Python3

# Python 3 program to count substrings with same
# first and last characters
 
# Function to count substrings with same first and
# last characters
def countSubstrs(str, i, j, n):
 
    # base cases
    if (n == 1):
        return 1
    if (n <= 0):
        return 0
 
    res = (countSubstrs(str, i + 1, j, n - 1)
        + countSubstrs(str, i, j - 1, n - 1)
        - countSubstrs(str, i + 1, j - 1, n - 2))    
 
    if (str[i] == str[j]):
        res += 1
 
    return res
 
# driver code
str = "abcab"
n = len(str)
print(countSubstrs(str, 0, n - 1, n))
 
# This code is contributed by Smitha

                    

Javascript

<script>
 
// Javascript program to count substrings
// with same first and last characters
 
// Function to count substrings
// with same first and
// last characters
function countSubstrs(str, i, j, n)
{
       
    // Base cases
    if (n == 1)
        return 1;
    if (n <= 0)
        return 0;
   
    let res = countSubstrs(str, i + 1, j, n - 1) +
              countSubstrs(str, i, j - 1, n - 1) -
          countSubstrs(str, i + 1, j - 1, n - 2);    
   
    if (str[i] == str[j])
        res++;
   
    return res;
}
 
// Driver code
let str = "abcab";
let n = str.length;
 
document.write(countSubstrs(str, 0, n - 1, n));
 
// This code is contributed by rameshtravel07
 
</script>

                    

C#

// C# program to count substrings
// with same first and last characters
using System;
 
class GFG {
     
    // Function to count substrings
    // with same first and
    // last characters
    static int countSubstrs(string str, int i,
                                 int j, int n)
    {
         
        // base cases
        if (n == 1)
            return 1;
        if (n <= 0)
            return 0;
     
        int res = countSubstrs(str, i + 1, j, n - 1)
                + countSubstrs(str, i, j - 1, n - 1)
            - countSubstrs(str, i + 1, j - 1, n - 2);    
     
        if (str[i] == str[j])
            res++;
     
        return res;
    }
     
    // Driver code
    public static void Main ()
    {
        string str = "abcab";
        int n = str.Length;
         
        Console.WriteLine(
                countSubstrs(str, 0, n - 1, n));
    }
}
 
// This code is contributed by vt_m.

                    

PHP

<?php
// PHP program to count
// substrings with same
// first and last characters
 
//Function to count substrings
// with same first and
// last characters
function countSubstrs($str, $i,
                      $j, $n)
{
     
    // base cases
    if ($n == 1)
        return 1;
    if ($n <= 0)
        return 0;
 
    $res = countSubstrs($str, $i + 1, $j, $n - 1) +
           countSubstrs($str, $i, $j - 1, $n - 1) -
           countSubstrs($str, $i + 1, $j - 1, $n - 2);    
 
    if ($str[$i] == $str[$j])
        $res++;
 
    return $res;
}
 
// Driver Code
$str = "abcab";
$n = strlen($str);
echo(countSubstrs($str, 0, $n - 1, $n));
 
// This code is contributed by Ajit.
?>

                    

Output
7


The time complexity of above solution is exponential. In Worst case, we may end up doing O(3n) operations.

Auxiliary Space:  O(n), where n is the length of string.

This is because when string is passed in the function it creates a copy of itself in stack.
 


There is also a divide and conquer recursive approach

The idea is to split the string in half until we get one element and have our base case return 2 things

  1. a map containing the character to the number of occurrences (i.e a:1 since its the base case)
  2. (This is the total number of substring with start and end with the same char. Since the base case is a string of size one, it starts and ends with the same character)

Then combine the solutions of the left and right division of the string, then return a new solution based on left and right. This new solution can be constructed by multiplying the common characters between the left and right set and adding to the solution count from left and right, and returning a map containing element occurrence count and solution count

Implementation:

Java

import java.util.HashMap;
import java.util.Map;
 
public class CountSubstr {
    public static void main(String[] args)
    {
        System.out.println(countSubstr("abcab"));
    }
 
    public static int countSubstr(String s)
    {
        if (s.length() == 0) {
            return 0;
        }
        Map<Character, Integer> charMap = new HashMap<>();
        int[] numSubstr = { 0 };
        countSubstrHelper(s, 0, s.length() - 1, charMap,
                          numSubstr);
        return numSubstr[0];
    }
 
    public static void
    countSubstrHelper(String string, int start, int end,
                      Map<Character, Integer> charMap,
                      int[] numSubstr)
    {
        if (start
            >= end) { // our base case for the recursion.
                      // When we have one character
            charMap.put(string.charAt(start), 1);
            numSubstr[0] = 1;
            return;
        }
        int mid = (start + end) / 2;
        Map<Character, Integer> mapLeft = new HashMap<>();
        int[] numSubstrLeft = { 0 };
        countSubstrHelper(
            string, start, mid, mapLeft,
            numSubstrLeft); // solve the left half
        Map<Character, Integer> mapRight = new HashMap<>();
        int[] numSubstrRight = { 0 };
        countSubstrHelper(
            string, mid + 1, end, mapRight,
            numSubstrRight); // solve the right half
        // add number of substrings from left and right
        numSubstr[0] = numSubstrLeft[0] + numSubstrRight[0];
 
        // multiply the characters from left set with
        // matching characters from right set then add to
        // total number of substrings
        for (char ch : mapLeft.keySet()) {
            if (mapRight.containsKey(ch)) {
                numSubstr[0]
                    += mapLeft.get(ch) * mapRight.get(ch);
            }
        }
 
        // Add all the key,value pairs from right map to
        // left map
        for (char ch : mapRight.keySet()) {
            if (mapLeft.containsKey(ch)) {
                mapLeft.put(ch, mapLeft.get(ch)
                                    + mapRight.get(ch));
            }
            else {
                mapLeft.put(ch, mapRight.get(ch));
            }
        }
        // Return the map of character and the sum of
        // substring from left, right and self
        charMap.putAll(mapLeft);
    }
}

                    

Python3

# code
def countSubstr(s):
    if len(s) == 0:
        return 0
    charMap, numSubstr = countSubstrHelper(s, 0, len(s)-1)
    return numSubstr
 
 
def countSubstrHelper(string, start, end):
    if start >= end:  # our base case for the recursion. When we have one character
        return {string[start]: 1}, 1
    mid = (start + end)//2
    mapLeft, numSubstrLeft = countSubstrHelper(
        string, start, mid)  # solve the left half
    mapRight, numSubstrRight = countSubstrHelper(
        string, mid+1, end)  # solve the right half
    # add number of substrings from left and right
    numSubstrSelf = numSubstrLeft + numSubstrRight
 
    # multiply the characters from left set with matching characters from right set
    # then add to total number of substrings
    for char in mapLeft:
        if char in mapRight:
            numSubstrSelf += mapLeft[char] * mapRight[char]
 
    # Add all the key,value pairs from right map to left map
    for char in mapRight:
        if char in mapLeft:
            mapLeft[char] += mapRight[char]
        else:
            mapLeft[char] = mapRight[char]
    # Return the map of character and the sum of substring from left, right and self
    return mapLeft, numSubstrSelf
 
 
print(countSubstr("abcab"))
 
# Contributed by Xavier Jean Baptiste

                    

Javascript

// JavaScript code
 
function countSubstr(s) {
  if (s.length == 0) {
    return 0;
  }
  let [charMap, numSubstr] = countSubstrHelper(s, 0, s.length - 1);
  return numSubstr;
}
 
function countSubstrHelper(string, start, end) {
 
   // our base case for the recursion. When we have one character
  if (start >= end) {
    return [{ [string[start]]: 1 }, 1];
  }
  let mid = Math.floor((start + end) / 2);
   
  // solve the left half
  let [mapLeft, numSubstrLeft] = countSubstrHelper(string, start, mid);
   
  // solve the right half
  // add number of substrings from left and right
  let [mapRight, numSubstrRight] = countSubstrHelper(string, mid + 1, end);
  let numSubstrSelf = numSubstrLeft + numSubstrRight;
   
    // multiply the characters from left set with matching characters from right set
    // then add to total number of substrings
  for (let char in mapLeft) {
    if (char in mapRight) {
      numSubstrSelf += mapLeft[char] * mapRight[char];
    }
  }
    
   // Add all the key,value pairs from right map to left map
  for (let char in mapRight) {
    if (char in mapLeft) {
      mapLeft[char] += mapRight[char];
    } else {
      mapLeft[char] = mapRight[char];
    }
  }
   
 // Return the map of character and the sum of substring from left, right and self
  return [mapLeft, numSubstrSelf];
}
 
console.log(countSubstr("abcab"));
 
// This code is contributed by adityashatmfh

                    

C#

using System;
using System.Collections.Generic;
 
public class CountSubstr {
    public static void Main(string[] args)
    {
        Console.WriteLine(countSubstr("abcab"));
    }
 
    public static int countSubstr(string s)
    {
        if (s.Length == 0) {
            return 0;
        }
        Dictionary<char, int> charMap
            = new Dictionary<char, int>();
        int[] numSubstr = { 0 };
        countSubstrHelper(s, 0, s.Length - 1, charMap,
                          numSubstr);
        return numSubstr[0];
    }
 
    public static void
    countSubstrHelper(string s, int start, int end,
                      Dictionary<char, int> charMap,
                      int[] numSubstr)
    {
        if (start >= end) {
            // our base case for the recursion. When we have
            // one character
            charMap[s[start]] = 1;
            numSubstr[0] = 1;
            return;
        }
        int mid = (start + end) / 2;
        Dictionary<char, int> mapLeft
            = new Dictionary<char, int>();
        int[] numSubstrLeft = { 0 };
        countSubstrHelper(
            s, start, mid, mapLeft,
            numSubstrLeft); // solve the left half
        Dictionary<char, int> mapRight
            = new Dictionary<char, int>();
        int[] numSubstrRight = { 0 };
        countSubstrHelper(
            s, mid + 1, end, mapRight,
            numSubstrRight); // solve the right half
        // add number of substrings from left and right
        numSubstr[0] = numSubstrLeft[0] + numSubstrRight[0];
 
        // multiply the characters from left set with
        // matching characters from right set then add to
        // total number of substrings
        foreach(char ch in mapLeft.Keys)
        {
            if (mapRight.ContainsKey(ch)) {
                numSubstr[0] += mapLeft[ch] * mapRight[ch];
            }
        }
 
        // Add all the key,value pairs from right map to
        // left map
        foreach(char ch in mapRight.Keys)
        {
            if (mapLeft.ContainsKey(ch)) {
                mapLeft[ch] = mapLeft[ch] + mapRight[ch];
            }
            else {
                mapLeft[ch] = mapRight[ch];
            }
        }
        // Return the map of character and the sum of
        // substring from left, right and self
        foreach(KeyValuePair<char, int> entry in mapLeft)
        {
            charMap[entry.Key] = entry.Value;
        }
    }
}
// This code in contributed by shiv1o43g

                    

C++

#include <iostream>
#include <unordered_map>
using namespace std;
 
void countSubstrHelper(string s, int start, int end,
                       unordered_map<char, int>& charMap,
                       int& numSubstr)
{
    if (start >= end) { // base case
        charMap[s[start]] = 1;
        numSubstr = 1;
        return;
    }
    int mid = (start + end) / 2;
    unordered_map<char, int> mapLeft, mapRight;
    int numSubstrLeft = 0, numSubstrRight = 0;
    countSubstrHelper(s, start, mid, mapLeft,
                      numSubstrLeft); // solve the left half
    countSubstrHelper(
        s, mid + 1, end, mapRight,
        numSubstrRight); // solve the right half
 
    // add number of substrings from left and right
    numSubstr = numSubstrLeft + numSubstrRight;
 
    // multiply the characters from left set with matching
    // characters from right set then add to total number of
    // substrings
    for (auto it = mapLeft.begin(); it != mapLeft.end();
         ++it) {
        if (mapRight.find(it->first) != mapRight.end()) {
            numSubstr += it->second * mapRight[it->first];
        }
    }
 
    // Add all the key,value pairs from right map to left
    // map
    for (auto it = mapRight.begin(); it != mapRight.end();
         ++it) {
        if (mapLeft.find(it->first) != mapLeft.end()) {
            mapLeft[it->first] += it->second;
        }
        else {
            mapLeft[it->first] = it->second;
        }
    }
 
    // Return the map of character and the sum of substring
    // from left, right and self
    charMap = mapLeft;
}
 
int countSubstr(string s)
{
    if (s.length() == 0) {
        return 0;
    }
    unordered_map<char, int> charMap;
    int numSubstr = 0;
    countSubstrHelper(s, 0, s.length() - 1, charMap,
                      numSubstr);
    return numSubstr;
}
 
int main()
{
    cout << countSubstr("abcab") << endl;
    return 0;
}
 
// This code is contributed by shivhack999

                    

Output
7

The time complexity of the above solution is O(nlogn) with space complexity O(n) which occurs if all elements are distinct

 



Last Updated : 04 Apr, 2023
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads