Given Q queries in the form of 2D array arr[][] whose every row consists of two numbers L and R which signifies the range [L, R], the task is to find the sum of all perfect squares lying in this range.
Examples:
Input: Q = 2, arr[][] = {{4, 9}, {4, 16}}
Output: 13 29
Explanation:
From 4 to 9: only 4 and 9 are perfect squares. Therefore, 4 + 9 = 13.
From 4 to 16: 4, 9 and 16 are the perfect squares. Therefore, 4 + 9 + 16 = 29.
Input: Q = 4, arr[][] = {{1, 10}, {1, 100}, {2, 25}, {4, 50}}
Output: 14 385 54 139
Approach: The idea is to use a prefix sum array. The sum all squares are precomputed and stored in an array pref[] so that every query can be answered in O(1) time. Every ‘i’th index in the pref[] array represents the sum of perfect squares from 1 to that number. Therefore, the sum of perfect squares from the given range ‘L’ to ‘R’ can be found as follows:
sum = pref[R] - pref[L - 1]
Below is the implementation of the above approach:
CPP
// C++ program to find the sum of all// perfect squares in the given range#include <bits/stdc++.h>#define ll intusing namespace std;// Array to precompute the sum of squares// from 1 to 100010 so that for every// query, the answer can be returned in O(1).long long pref[100010];// Function to check if a number is// a perfect square or notint isPerfectSquare(long long int x){ // Find floating point value of // square root of x. long double sr = sqrt(x); // If square root is an integer return ((sr - floor(sr)) == 0) ? x : 0;}// Function to precompute the perfect// squares upto 100000.void compute(){ for (int i = 1; i <= 100000; ++i) { pref[i] = pref[i - 1] + isPerfectSquare(i); }}// Function to print the sum for each queryvoid printSum(int L, int R){ int sum = pref[R] - pref[L - 1]; cout << sum << " ";}// Driver codeint main(){ // To calculate the precompute function compute(); int Q = 4; int arr[][2] = { { 1, 10 }, { 1, 100 }, { 2, 25 }, { 4, 50 } }; // Calling the printSum function // for every query for (int i = 0; i < Q; i++) { printSum(arr[i][0], arr[i][1]); } return 0;} |
Java
// Java program to find the sum of all// perfect squares in the given rangeclass GFG{// Array to precompute the sum of squares// from 1 to 100010 so that for every// query, the answer can be returned in O(1).static int []pref = new int[100010];// Function to check if a number is// a perfect square or notstatic int isPerfectSquare(int x){ // Find floating point value of // square root of x. double sr = Math.sqrt(x); // If square root is an integer return ((sr - Math.floor(sr)) == 0) ? x : 0;}// Function to precompute the perfect// squares upto 100000.static void compute(){ for (int i = 1; i <= 100000; ++i) { pref[i] = pref[i - 1] + isPerfectSquare(i); }}// Function to print the sum for each querystatic void printSum(int L, int R){ int sum = pref[R] - pref[L - 1]; System.out.print(sum+ " ");}// Driver codepublic static void main(String[] args){ // To calculate the precompute function compute(); int Q = 4; int arr[][] = { { 1, 10 }, { 1, 100 }, { 2, 25 }, { 4, 50 } }; // Calling the printSum function // for every query for (int i = 0; i < Q; i++) { printSum(arr[i][0], arr[i][1]); }}}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 program to find the sum of all # perfect squares in the given range from math import sqrt, floor# Array to precompute the sum of squares # from 1 to 100010 so that for every # query, the answer can be returned in O(1). pref = [0]*100010; # Function to check if a number is # a perfect square or not def isPerfectSquare(x) : # Find floating point value of # square root of x. sr = sqrt(x); # If square root is an integer rslt = x if (sr - floor(sr) == 0) else 0; return rslt; # Function to precompute the perfect # squares upto 100000. def compute() : for i in range(1 , 100001) : pref[i] = pref[i - 1] + isPerfectSquare(i); # Function to print the sum for each query def printSum( L, R) : sum = pref[R] - pref[L - 1]; print(sum ,end= " "); # Driver code if __name__ == "__main__" : # To calculate the precompute function compute(); Q = 4; arr = [ [ 1, 10 ], [ 1, 100 ], [ 2, 25 ], [ 4, 50 ] ]; # Calling the printSum function # for every query for i in range(Q) : printSum(arr[i][0], arr[i][1]); # This code is contributed by AnkitRai01 |
C#
// C# program to find the sum of all// perfect squares in the given rangeusing System;class GFG{// Array to precompute the sum of squares// from 1 to 100010 so that for every// query, the answer can be returned in O(1).static int []pref = new int[100010];// Function to check if a number is// a perfect square or notstatic int isPerfectSquare(int x){ // Find floating point value of // square root of x. double sr = Math.Sqrt(x); // If square root is an integer return ((sr - Math.Floor(sr)) == 0) ? x : 0;}// Function to precompute the perfect// squares upto 100000.static void compute(){ for (int i = 1; i <= 100000; ++i) { pref[i] = pref[i - 1] + isPerfectSquare(i); }}// Function to print the sum for each querystatic void printSum(int L, int R){ int sum = pref[R] - pref[L - 1]; Console.Write(sum+ " ");}// Driver codepublic static void Main(String[] args){ // To calculate the precompute function compute(); int Q = 4; int [,]arr = { { 1, 10 }, { 1, 100 }, { 2, 25 }, { 4, 50 } }; // Calling the printSum function // for every query for (int i = 0; i < Q; i++) { printSum(arr[i, 0], arr[i, 1]); }}}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// Javascript program to find the sum of all// perfect squares in the given range// Array to precompute the sum of squares// from 1 to 100010 so that for every// query, the answer can be returned in O(1).var pref= Array(100010).fill(0);// Function to check if a number is// a perfect square or notfunction isPerfectSquare(x){ // Find floating point value of // square root of x. var sr = Math.sqrt(x); // If square root is an integer return ((sr - Math.floor(sr)) == 0) ? x : 0;}// Function to precompute the perfect// squares upto 100000.function compute(){ for (var i = 1; i <= 100000; ++i) { pref[i] = pref[i - 1] + isPerfectSquare(i); }}// Function to print the sum for each queryfunction printSum(L, R){ var sum = pref[R] - pref[L - 1]; document.write(sum + " ");}// Driver code// To calculate the precompute functioncompute();var Q = 4;arr = [ [ 1, 10 ], [ 1, 100 ], [ 2, 25 ], [ 4, 50 ] ]; // Calling the printSum function// for every queryfor(var i = 0; i < Q; i++) printSum(arr[i][0], arr[i][1]);</script> |
Output:
14 385 54 139
Time Complexity: O(Q + 10000 * x)
Auxiliary Space: O(100010)
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