You are given an array of strings str[], the task is to find the score of a given string s from the array. The score of a string is defined as the product of the sum of its characters’ alphabetical values with the position of the string in the array.
Examples:
Input: str[] = {“sahil”, “shashanak”, “sanjit”, “abhinav”, “mohit”}, s = “abhinav”
Output: 228
Sum of alphabetical values of “abhinav” = 1 + 2 + 8 + 9 + 14 + 1 + 22 = 57
Position of “abhinav” in str is 4, 57 x 4 = 228Input: str[] = {“neveropen”, “algorithms”, “stack”}, s = “algorithms”
Output: 244
Naive approach:
- Find the sum of characters of given string s and store it into variable ‘a’
- Find the position of given string s in str and store it into variable ‘b’
- Find the product of a * b.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the required string scoreint strScore(string str[], string s, int n){ int a = 0; // Find the sum of character of given string s // and store into variable a for (auto c : s) a += (c - 'a') + 1; int b = 0; // Find the position of given string s in str // and store into variable b for (int i = 0; i < n; i++) { if (str[i] == s) { b = i + 1; break; } } // Find the product of a * b. return a * b;}// Driver codeint main(){ string str[] = { "sahil", "shashanak", "sanjit", "abhinav", "mohit" }; string s = "abhinav"; int n = sizeof(str) / sizeof(str[0]); int score = strScore(str, s, n); cout << score; return 0;} |
Java
// Java implementation of the approachimport java.util.*;public class GFG { // Function to return the required string score static int strScore(String str[], String s, int n) { int a = 0; // Find the sum of character of given string s // and store into variable a for (char c : s.toCharArray()) a += (c - 'a') + 1; int b = 0; // Find the position of given string s in str // and store into variable b for (int i = 0; i < n; i++) { if (str[i].equals(s)) { b = i + 1; break; } } // Find the product of a * b. return a * b; } // Driver code public static void main(String[] args) { String str[] = { "sahil", "shashanak", "sanjit", "abhinav", "mohit" }; String s = "abhinav"; int n = str.length; int score = strScore(str, s, n); System.out.println(score); }}// This code is contributed by Karandeep1234 |
Python3
# Python implementation of the approach# Function to return the required string scoredef strScore(str, s, n): a = 0; # Find the sum of character of given string s # and store into variable a for i in range(0,len(s)): a += (ord(s[i]) - ord('a')) + 1; b = 0; # Find the position of given string s in str # and store into variable b for i in range(0,n): if (str[i] == s): b = i + 1; break; # Find the product of a * b. return a * b;# Driver codestr = [ "sahil", "shashanak", "sanjit", "abhinav", "mohit" ];s = "abhinav";n = len(str);score = strScore(str, s, n);print(score);# This code is contributed by poojaagarwal2. |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;class GfG { // Function to return the required string score static int strScore(String[] str, String s, int n) { int a = 0; // Find the sum of character of given string s // and store into variable a for (int i = 0; i < s.Length; i++) a += (s[i] - 'a') + 1; // Console.WriteLine(a); int b = 0; // Find the position of given string s in str // and store into variable b for (int i = 0; i < n; i++) { if (str[i] == s) { b = i + 1; break; } } // Find the product of a * b. return a * b; } // Driver code public static void Main() { string[] str = { "sahil", "shashanak", "sanjit", "abhinav", "mohit" }; string s = "abhinav"; int n = str.Length; Console.WriteLine(strScore(str, s, n)); }} |
Javascript
// Javascript implementation of the approach// Function to return the required string scorefunction strScore(str, s, n){ let a = 0; // Find the sum of character of given string s // and store into variable a for (let i=0; i<s.length; i++) { a += s[i].charCodeAt(0)-'a'.charCodeAt(0) + 1; } let b = 0; // Find the position of given string s in str // and store into variable b for (let i = 0; i < n; i++) { if (str[i] == s) { b = i + 1; break; } } // Find the product of a * b. return a * b;}// Driver code let str = [ "sahil", "shashanak", "sanjit", "abhinav", "mohit" ]; let s = "abhinav"; let n = str.length; let score = strScore(str, s, n); document.write(score); |
Output
228
Time Complexity: O(max(N, M)), where N and M are the lengths of given string ‘str’ and the length of string ‘s’
Auxiliary Space: O(1)
Approach:
In SET 1, we saw an approach where every time a query is being executed, the position of the string has to be found with a single traversal of str[]. This can be optimized when there are a number of queries using a hash table.
- Create a hash map of all the strings present in str[] along with their respective positions in the array.
- Then for every query s, check if s is present in the map. If yes then calculate the sum of the alphabetical values of s and store it in sum.
- Print sum * pos where pos is the value associated with s in map i.e. its position in str[].
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the required string scoreint strScore(string str[], string s, int n){ // create a hash map of strings in str unordered_map<string, int> m; // Store every string in the map // along with its position in the array for (int i = 0; i < n; i++) m[str[i]] = i + 1; // If given string is not present in str[] if (m.find(s) == m.end()) return 0; int score = 0; for (int i = 0; i < s.length(); i++) score += s[i] - 'a' + 1; // Multiply sum of alphabets with position score = score * m[s]; return score;}// Driver codeint main(){ string str[] = { "neveropen", "algorithms", "stack" }; string s = "algorithms"; int n = sizeof(str) / sizeof(str[0]); int score = strScore(str, s, n); cout << score; return 0;} |
Java
// Java implementation of the approachimport java.util.HashMap;import java.util.Map;class GfG{ // Function to return the required string score static int strScore(String str[], String s, int n) { // create a hash map of strings in str HashMap<String, Integer> m = new HashMap<>(); // Store every string in the map // along with its position in the array for (int i = 0; i < n; i++) m.put(str[i], i + 1); // If given string is not present in str[] if (!m.containsKey(s)) return 0; int score = 0; for (int i = 0; i < s.length(); i++) score += s.charAt(i) - 'a' + 1; // Multiply sum of alphabets with position score = score * m.get(s); return score; } // Driver code public static void main(String []args) { String str[] = { "neveropen", "algorithms", "stack" }; String s = "algorithms"; int n = str.length; System.out.println(strScore(str, s, n)); }}// This code is contributed by Rituraj Jain |
Python3
# Python3 implementation of the approach # Function to return the required # string score def strScore(string, s, n) : # create a hash map of strings in str m = {} # Store every string in the map # along with its position in the array for i in range(n) : m[string[i]] = i + 1 # If given string is not present in str[] if s not in m.keys() : return 0 score = 0 for i in range(len(s)) : score += ord(s[i]) - ord('a') + 1 # Multiply sum of alphabets # with position score = score * m[s] return score # Driver code if __name__ == "__main__" : string = [ "neveropen", "algorithms", "stack" ] s = "algorithms" n = len(string) score = strScore(string, s, n); print(score) # This code is contributed by Ryuga |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;class GfG{ // Function to return the required string score static int strScore(string [] str, string s, int n) { // create a hash map of strings in str Dictionary<string, int> m = new Dictionary<string, int>(); // Store every string in the map // along with its position in the array for (int i = 0; i < n; i++) m[str[i]] = i + 1; // If given string is not present in str[] if (!m.ContainsKey(s)) return 0; int score = 0; for (int i = 0; i < s.Length; i++) score += s[i] - 'a' + 1; // Multiply sum of alphabets with position score = score * m[s]; return score; } // Driver code public static void Main() { string [] str = { "neveropen", "algorithms", "stack" }; string s = "algorithms"; int n = str.Length; Console.WriteLine(strScore(str, s, n)); }}// This code is contributed by ihritik |
Javascript
<script>// JavaScript Program to implement// the above approach // Function to return the required string score function strScore(str, s, n) { // create a hash map of strings in str let m = new Map(); // Store every string in the map // along with its position in the array for (let i = 0; i < n; i++) m.set(str[i], i + 1); // If given string is not present in str[] if (!m.has(s)) return 0; let score = 0; for (let i = 0; i < s.length; i++) score += s[i].charCodeAt() - 'a'.charCodeAt() + 1; // Multiply sum of alphabets with position score = score * m.get(s); return score; } // Driver Code let str = [ "neveropen", "algorithms", "stack" ]; let s = "algorithms"; let n = str.length; document.write(strScore(str, s, n));</script> |
Output
244
Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(n), for using map
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