Recursive solution to count substrings with same first and last characters
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.
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 codeint 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 charactersclass 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 charactersdef 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 codestr = "abcab"n = len(str)print(countSubstrs(str, 0, n - 1, n))# This code is contributed by Smitha |
C#
// C# program to count substrings// with same first and last charactersusing 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 charactersfunction 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.?> |
Javascript
<script>// Javascript program to count substrings// with same first and last characters// Function to count substrings// with same first and// last charactersfunction 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 codelet str = "abcab";let n = str.length;document.write(countSubstrs(str, 0, n - 1, n));// This code is contributed by rameshtravel07</script> |
7
The time complexity of above solution is exponential. In Worst case, we may end up doing O(3n) operations.
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 – 1 (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
Python3
# codedef countSubstr(s): if len(s) == 0: return 0 charMap, numSubstr = countSubstrHelper(s, 0, len(s)-1) return numSubstrdef 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, numSubstrSelfprint(countSubstr("abcab"))# Contributed by Xavier Jean Baptiste |
7
The time complexity of the above solution is O(nlogn) with space complexity O(n) which occurs if all elements are distinct
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