Hash Table
We use the hash table to keep the track of the indices of each character in the string.For each character,iterate over its indices
and calculate the sum of indices between each pair of indices. Last add these sums to the total sum of distances.
Steps:
- First, create an unordered map to store the indices of each character in the input string S.
- Then iterate over each character in the map.
- And for each character, iterate over its indices.
- Next, calculate the sum of distances between each pair of indices.
- add these sums to the total sum of distances.
- Finally, return the total sum of distances.
Below is the code implementation for the above approach:
C++
// C++ program #include <iostream> #include <unordered_map> #include <vector> #include <cmath> using namespace std; int sum_of_distances(string S) { int n = S.length(); // create a hash table to store indices unordered_map< char , vector< int >> indices; int total_distance = 0; for ( int i = 0; i < n; i++) { indices[S[i]].push_back(i); } for ( auto & indices_list : indices) { int m = indices_list.second.size(); for ( int i = 0; i < m; i++) { for ( int j = i+1; j < m; j++) { total_distance += abs (indices_list.second[j] - indices_list.second[i]); } } } // return the total sum of distances return total_distance; } // Driver Code int main() { string S = "ababa" ; cout << sum_of_distances(S) << endl; return 0; } |
Java
import java.util.ArrayList; import java.util.HashMap; import java.util.List; import java.util.Map; public class GFG { public static int sumOfDistances(String S) { int n = S.length(); // Create a HashMap to store indices Map<Character, List<Integer>> indices = new HashMap<>(); int totalDistance = 0 ; // Populate the HashMap with indices of each character in the string for ( int i = 0 ; i < n; i++) { char ch = S.charAt(i); if (!indices.containsKey(ch)) { indices.put(ch, new ArrayList<>()); } indices.get(ch).add(i); } // Calculate the sum of distances between all pairs of indices for each character for (Map.Entry<Character, List<Integer>> entry : indices.entrySet()) { List<Integer> indicesList = entry.getValue(); int m = indicesList.size(); for ( int i = 0 ; i < m; i++) { for ( int j = i + 1 ; j < m; j++) { totalDistance += Math.abs(indicesList.get(j) - indicesList.get(i)); } } } // Return the total sum of distances return totalDistance; } public static void main(String[] args) { String S = "ababa" ; System.out.println(sumOfDistances(S)); } } |
Python3
def sum_of_distances(S): n = len (S) # Create a dictionary to store indices indices = {} total_distance = 0 for i in range (n): if S[i] not in indices: indices[S[i]] = [i] else : indices[S[i]].append(i) for indices_list in indices.values(): m = len (indices_list) for i in range (m): for j in range (i + 1 , m): total_distance + = abs (indices_list[j] - indices_list[i]) # Return the total sum of distances return total_distance # Driver Code if __name__ = = "__main__" : S = "ababa" print (sum_of_distances(S)) |
C#
using System; using System.Collections.Generic; class Program { static int SumOfDistances( string S) { int n = S.Length; // Create a dictionary to store indices Dictionary< char , List< int >> indices = new Dictionary< char , List< int >>(); int totalDistance = 0; // Populate the dictionary with character indices for ( int i = 0; i < n; i++) { char c = S[i]; if (!indices.ContainsKey(c)) { indices = new List< int >(); } indices.Add(i); } // Calculate the total sum of distances foreach ( var indicesList in indices.Values) { int m = indicesList.Count; for ( int i = 0; i < m; i++) { for ( int j = i + 1; j < m; j++) { totalDistance += Math.Abs(indicesList[j] - indicesList[i]); } } } // Return the total sum of distances return totalDistance; } static void Main() { string S = "ababa" ; Console.WriteLine(SumOfDistances(S)); } } |
Javascript
function sumOfDistances(S) { const n = S.length; // Create an object to store indices const indices = {}; let totalDistance = 0; // Populate the indices object for (let i = 0; i < n; i++) { const char = S[i]; if (indices[char]) { indices[char].push(i); } else { indices[char] = [i]; } } for (const indicesList of Object.values(indices)) { const m = indicesList.length; for (let i = 0; i < m; i++) { for (let j = i + 1; j < m; j++) { totalDistance += Math.abs(indicesList[j] - indicesList[i]); } } } // Return the total sum of distances return totalDistance; } // Driver Code const S = "ababa" ; console.log(sumOfDistances(S)); |
10
Time Complexity: O(n^2), where n is the length of the input string S.
Auxiliary Space: O(n), where n is the length of the input string S.
Sum of all distances between occurrences of same characters in a given string
Given a string S, the task is to find the sum of distances between all pairs of indices from the given string which contains the same character.
Examples:
Input: S = “ababa”
Output: 10
Explanation:
The pair of indices having same character are: (0, 2) (0, 4) (1, 3) (2, 4)
Sum of absolute differences between these pair of indices = |2 – 0| + |4 – 0| + |1 – 3| + |2 – 4| = 10.
Therefore, the required answer is 10.Input: S = “ttt”
Output: 4
Naive Approach: The simplest approach to solve the problem is to traverse the string and for each character encountered, traverse the remaining string on its right to find occurrences of that character. For every repetition of characters found, keep adding the absolute difference between the concerned indices to the answer. Finally, print the sum obtained.
Time Complexity: O(N2)
Auxiliary Space: O(1)
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to calculate the sum // of distances between occurrences // of same characters in a string int findSum(string s) { int sum = 0; for ( int i = 0; i < s.size(); i++) { for ( int j = i + 1; j < s.size(); j++) { // If similar characters are found if (s[i] == s[j]) { // Add the difference // of their positions sum += (j - i); } } } // Return the answer return sum; } // Driver Code int main() { string s = "ttt" ; cout << findSum(s) << endl; } |
Java
// Java program for the above approach import java.util.*; class GFG{ // Function to calculate the sum // of distances between occurrences // of same characters in a String static int findSum(String s) { int sum = 0 ; for ( int i = 0 ; i < s.length(); i++) { for ( int j = i + 1 ; j < s.length(); j++) { // If similar characters are found if (s.charAt(i) == s.charAt(j)) { // Add the difference // of their positions sum += (j - i); } } } // Return the answer return sum; } // Driver Code public static void main(String[] args) { String s = "ttt" ; System.out.print(findSum(s) + "\n" ); } } // This code is contributed by shikhasingrajput |
Python3
# Python3 program for the above approach # Function to calculate the sum # of distances between occurrences # of same characters in a string def findSum(s): sum = 0 for i in range ( len (s)): for j in range (i + 1 , len (s)): # If similar characters are found if (s[i] = = s[j]): # Add the difference # of their positions sum + = (j - i) # Return the answer return sum # Driver Code s = "ttt" print (findSum(s)) # This code is contributed by code_hunt |
C#
// C# program for // the above approach using System; class GFG{ // Function to calculate the sum // of distances between occurrences // of same characters in a String static int findSum(String s) { int sum = 0; for ( int i = 0; i < s.Length; i++) { for ( int j = i + 1; j < s.Length; j++) { // If similar characters // are found if (s[i] == s[j]) { // Add the difference // of their positions sum += (j - i); } } } // Return the answer return sum; } // Driver Code public static void Main(String[] args) { String s = "ttt" ; Console.Write(findSum(s) + "\n" ); } } // This code is contributed by shikhasingrajput |
Javascript
<script> // javascript program for the // above approach // Function to calculate the sum // of distances between occurrences // of same characters in a String function findSum(s) { let sum = 0; for (let i = 0; i < s.length; i++) { for (let j = i + 1; j < s.length; j++) { // If similar characters are found if (s[i] == s[j]) { // Add the difference // of their positions sum += (j - i); } } } // Return the answer return sum; } // Driver Code let s = "ttt" ; document.write(findSum(s) + "<br/>" ); // This code is contributed by target_2. </script> |
4
Efficient Approach: The above approach can be optimized based on the following observations:
- Initially for every character, assume that all its similar characters are at index 0.
- With the above assumption, the required sum becomes equal to:
Number of previously visited similar characters * Index of the character – sum of distances of those similar characters from index 0
Therefore, follow the steps below to solve the problem:
- Initialize two arrays visited[] and distance[] to store the frequency of each character present in the string and the distance between the previous occurrences of each character respectively.
- Traverse the string and for every character encountered, i.e. S[i], update the following:
- Add visited[S[i] * i – distance[S[i]] to the required sum.
- Increment visited[S[i]] to increase frequency of characters.
- Increase distance[S[i]] by i, to increase the distance from the previous occurrence of S[i], considered to be 0.
- Once the above steps are completed, print the sum obtained.
Below is the implementation of the above approach :
C++
#include <bits/stdc++.h> using namespace std; // Function to calculate the sum // of distances between occurrences // of same characters in a string int findSum(string s) { // Initially make all the distances // and number of characters visited as 0 vector< int > visited(256, 0); vector< int > distance(256, 0); int sum = 0; for ( int i = 0; i < s.length(); i++) { // Assuming that all the similar // characters are located at index 0 // Add visited[s[i]]*i to sum // and subtract the distances of // characters from index 0 sum += visited[s[i] - 'a' ] * i - distance[s[i] - 'a' ]; // Increment the number of // visited characters visited[s[i] - 'a' ]++; // Add the distance of the // character from position 0 // i.e., (i - 0) = i distance[s[i] - 'a' ] += i; } // Return the answer return sum; } int main() { string s = "ttt" ; cout << findSum(s) << "\n" ; return 0; } |
Java
// Java program for the above approach import java.io.*; class GFG{ // Function to calculate the sum // of distances between occurrences // of same characters in a string static int findSum(String s) { int [] visited = new int [ 256 ]; int [] distance = new int [ 256 ]; // Initially make all the distances // and number of characters visited as 0 for ( int i = 0 ; i < 256 ; i++) { visited[i] = 0 ; distance[i] = 0 ; } int sum = 0 ; for ( int i = 0 ; i < s.length(); i++) { // Assuming that all the similar // characters are located at index 0 // Add visited[s[i]]*i to sum // and subtract the distances of // characters from index 0 sum += visited[s.charAt(i)] * i - distance[s.charAt(i)]; // Increment the number of // visited characters visited[s.charAt(i)]++; // Add the distance of the // character from position 0 // i.e., (i - 0) = i distance[s.charAt(i)] += i; } // Return the answer return sum; } // Driver code public static void main (String[] args) { String s = "ttt" ; // Function call System.out.println(findSum(s)); } } // This code is contributed by offbeat |
Python3
# Python3 program for the above approach # Function to calculate the sum # of distances between occurrences # of same characters in a string def findSum(s): visited = [ 0 for i in range ( 256 )]; distance = [ 0 for i in range ( 256 )]; # Initially make all # the distances and number # of characters visited as 0 for i in range ( 256 ): visited[i] = 0 ; distance[i] = 0 ; sum = 0 ; for i in range ( len (s)): # Assuming that all the similar # characters are located at index 0 # Add visited[s[i]]*i to sum # and subtract the distances of # characters from index 0 sum + = visited[ ord (s[i])] * i - distance[ ord (s[i])]; # Increment the number of # visited characters visited[ ord (s[i])] + = 1 ; # Add the distance of the # character from position 0 # i.e., (i - 0) = i distance[ ord (s[i])] + = i; # Return the answer return sum ; # Driver code if __name__ = = '__main__' : s = "ttt" ; # Function call print (findSum(s)); # This code is contributed by Rajput-Ji |
C#
// C# program for the above approach using System; class GFG{ // Function to calculate the sum // of distances between occurrences // of same characters in a string static int findSum(String s) { int [] visited = new int [256]; int [] distance = new int [256]; // Initially make all the distances // and number of characters visited as 0 for ( int i = 0; i < 256; i++) { visited[i] = 0; distance[i] = 0; } int sum = 0; for ( int i = 0; i < s.Length; i++) { // Assuming that all the similar // characters are located at index 0 // Add visited[s[i]]*i to sum // and subtract the distances of // characters from index 0 sum += visited[s[i]] * i - distance[s[i]]; // Increment the number of // visited characters visited[s[i]]++; // Add the distance of the // character from position 0 // i.e., (i - 0) = i distance[s[i]] += i; } // Return the answer return sum; } // Driver code public static void Main(String[] args) { String s = "ttt" ; // Function call Console.WriteLine(findSum(s)); } } // This code is contributed by Amit Katiyar |
Javascript
<script> // javascript program for the above approach // Function to calculate the sum // of distances between occurrences // of same characters in a string function findSum(s) { var visited = Array(256).fill(0); var distance = Array(256).fill(0); var sum = 0; var i; for (i = 0; i < s.length; i++) { // Assuming that all the similar // characters are located at index 0 // Add visited[s[i]]*i to sum // and subtract the distances of // characters from index 0 sum += visited[s.charCodeAt(i)] * i - distance[s.charCodeAt(i)]; // Increment the number of // visited characters visited[s.charCodeAt(i)]++; // Add the distance of the // character from position 0 // i.e., (i - 0) = i distance[s.charCodeAt(i)] += i; } // Return the answer return sum; } // Driver code var s = "ttt" ; // Function call document.write(findSum(s)); </script> |
4
Time Complexity: O(N)
Auxiliary Space: O(1)