Given string str, the task is to check the sum of ASCII values of all characters in this string is a perfect cube or not.
Examples :
Input: str = “ll”
Output: Yes
ASCII value of l = 108
Therefore, sum of ASCII values = 108 + 108 = 216
which is a perfect cube 6 (6 * 6 * 6 = 216)Input: str = “a”
Output: No
ASCII value of a = 97
Therefore, sum of ASCII values = 97
which is not a perfect cube
Algorithm
- For each character in the String, find out its ASCII value
- Calculate sum of ASCII values of all characters
- Check whether this sum is a perfect cube or not.
- If the sum is a perfect cube, then print “Yes” otherwise “No”
Below is the implementation of the above approach:
C++
// C++ program to find if string is a// perfect cube or not.#include <bits/stdc++.h>using namespace std;bool isPerfectCubeString(string str){ int sum = 0; // Finding ASCII values of each // character and finding its sum for (int i = 0; i < str.length(); i++) sum += (int)str[i]; // Find the cube root of sum long double cr = round(cbrt(sum)); // Check if sum is a perfect cube return (cr * cr * cr == sum);}// Driver codeint main(){ string str = "ll"; if (isPerfectCubeString(str)) cout << "Yes"; else cout << "No";} |
Java
// Java program to find if String is a// perfect cube or not.import java.util.*;class GFG{ static boolean isPerfectCubeString(String str){ int sum = 0; // Finding ASCII values of each // character and finding its sum for (int i = 0; i < str.length(); i++) sum += (int)str.charAt(i); // Find the cube root of sum double cr = Math.round(Math.cbrt(sum)); // Check if sum is a perfect cube return (cr * cr * cr == sum);} // Driver codepublic static void main(String[] args){ String str = "ll"; if (isPerfectCubeString(str)) System.out.print("Yes"); else System.out.print("No");}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program to find if string is a# perfect cube or not.from math import ceildef isPerfectCubeString(str1): sum = 0 # Finding ASCII values of each # character and finding its sum for i in range(len(str1)): sum += ord(str1[i]) # Find the cube root of sum cr = ceil((sum)**(1/3)) # Check if sum is a perfect cube return (cr * cr * cr == sum)# Driver codestr1 = "ll"if (isPerfectCubeString(str1)): print("Yes")else: print("No")# This code is contributed by mohit kumar 29 |
C#
// C# program to find if String is a// perfect cube or not.using System;class GFG{static bool isPerfectCubeString(String str){ int sum = 0; // Finding ASCII values of each // character and finding its sum for (int i = 0; i < str.Length; i++) sum += (int)str[i]; // Find the cube root of sum double cr = Math.Round(Math.Pow(sum, (double) 1 / 3)); // Check if sum is a perfect cube return (cr * cr * cr == sum);}// Driver codepublic static void Main(String[] args){ String str = "ll"; if (isPerfectCubeString(str)) Console.Write("Yes"); else Console.Write("No");}}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// JavaScript program to find if string// is a perfect cube or not.function isPerfectCubeString(str){ var sum = 0; // Finding ASCII values of each // character and finding its sum for(var i = 0; i < str.length; i++) { sum += str.charCodeAt(i); } // Find the cube root of sum var cr = Math.round(Math.cbrt(sum)); // Check if sum is a perfect cube return cr * cr * cr == sum;}// Driver codevar str = "ll";if (isPerfectCubeString(str)) document.write("Yes");else document.write("No");// This code is contributed by rdtank </script> |
Output:
Yes
Time complexity: O(n) where n is the length of the given string
Auxiliary space: O(1)
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