Consider the following snippet of Java code:int sumHelper(int n int a) {}'if (n= 0) return a;Belse return sumHelper(n-1, a + n*n);int sumSqr(int n) { return sumHelper(n, 0); }Your Task: Prove that for n ≥ 1, sumSqr(n) → 1² +2² + 3² + ... + n². To prove this, use induction toshow that sumHelper computes the “right thing." (Hint: How did we prove fact_helper in class?)

To prove that the function `sumSqr(n)` computes the sum of squares (12+22+32+⋯+n2) for (n≥1), let's…

Answered: Consider the following snippet of Java…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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%3B for ( int i = 0; i < n; ++i ) { for ( int j= 0; j< n; ++j) { sum += f( j, 10 ); } } iii. int sum = 0; %3D Assume f(n, m) runs in O( n + m) time.
int sum, k, i, j; int x[4] [4]={1,2,3,4}, {5,6,7,8},{9,8,7,3},{2, 1,7,1}; sum=x[0] [0]; for (k=1; k<=3;k++) sum+=x[k] [k]; Give the value in sum after the statements are executed:
bool isprime(long n) /* fixed from to https://www.geeksforgeeks.org/euclid-euler-theorem/?ref=lbp */{ // check whether a number is prime or not int i; for (i = 2; i * i <= n; i++) if (n % i == 0) return false; return true;}
sum= 0; for (int i = 0; i 1) { sum++; i= 1/2; } = 2*log2 (n) We denote by Ta(n), Tb (n), Te(n) the running time of the three fragments. 1. Give evaluations for Ta(n), Tb (n), Te(n). 2. Is T(n) = O(Ta(n)) ? Answer YES or NO and justify your answer. 3. Is Te(n) = (Ta(n)) ? Answer YES or NO and justify your answer.
Java:
solve in java Integer arrays originalArr and filterArr are read from input, each containing two elements. If an element in originalArr is not a multiple of the corresponding element in filterArr, replace the element in originalArr with the corresponding element in filterArr times 2. Ex: If the input is: 16 9 8 4 then the output is: 16 8 Note: Given integers A and B, A is a multiple of B only if A % B == 0. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 import java.util.Scanner; public class ReplaceNumbers { publicstaticvoidmain(String[] args) { Scannerscnr=newScanner(System.in); finalintNUM_VALS=2; int[] originalArr=newint[NUM_VALS]; int[] filterArr=newint[NUM_VALS]; inti; for (i=0; i<NUM_VALS; ++i) { originalArr[i] =scnr.nextInt(); } for (i=0; i<NUM_VALS; ++i) { filterArr[i] =scnr.nextInt(); } /* Your code goes here */ for (i = 0; i < originalArr.length; ++i) { System.out.print(originalArr[i] + " "); } System.out.println();…
Consider the sequence of numbers W,, W,, W2, W. defined recursively: W, = 1, W, = 2, W3 = 3, W,=W-3 *(2+Wn-1- W-a), for n 2 4. and the following recursive Java method for computing the number W.: public static int recW(int n) { if (n <= 0 ) { return 0; } else if (n <= 3) { return n; } else { return ( recW(n-3) * (2 * recW(n-1) - recW(n-3)) ); The method recW has an obvious inefficiency, which can be easily removed in the following way. O Compute first recW(n-3) and then use the computed value in the expression in the return statement. Convert the "if-else if-else" statements into a "switch" statement. Remove the unnecessary case of a non-positive argument n. O Consider separately the three base cases n = 1, n = 2 and n = 3. O Compute first recW(n-1) and then use the computed value in the expression in the return statement.
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Please write the code for the given question in Java language.
int countTripletSumPermutations(int size , int* arr, int tripletSum){int count = 0 ; for(int i = 0 ; i < size - 2; i++) { if (tripletSum % arr[i] == 0) { for (int j =0 ; j < size - 1; j++) { if (tripletSum % (arr[i] * arr[j]) == 0) { int value = tripletSum /(arr[i] * arr[j]); for(int k = j + 1 ; k < size ; k++ ) if (arr[k] == value) count++ ; } } }} return count;}
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