del(int n){ j=(n/3)-1 while (n>0) for(i=n to i=j*3){ print "i"; 5. n--; }//end for 7 j=j/3; }//endwhile }//del() what is the asymptotic run time of del? anyalse each line and what is teh final run time is teh answer O(n^2/3)?
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del(int n){
j=(n/3)-1
while (n>0)
for(i=n to i=j*3){
print "i"; 5. n--;
}//end for 7 j=j/3;
}//endwhile
}//del()
what is the asymptotic run time of del?
anyalse each line and what is teh final run time
is teh answer O(n^2/3)?
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Solved in 2 steps
- del(int n){ j=(n/3)-1 while (n>0) for(i=n to i=j*3){ print "i"; 5. n--; }//end for 7 j=j/3; }//endwhile }//del() what is the asymptotic run time of del? anyalse each line and what is the final run timevoid deleteRange( int from, int to) {int i, j = 0;for (i = 0; i < counter; i++) {if (i <= from - 1 || i >= to + 1) {A[j] = A[i];j++;}}for (int i = 0; i < j; i++)cout << A[i] << " ";} Above method deletes range of elements from an array. Consider array A[] globally declared and counter is its size. Please explain the logic of above code in simple english (algorithm and comments).void deleteRange( int from, int to) { int i, j = 0; for (i = 0; i < counter; i++) { if (i <= from - 1 || i >= to + 1) { A[j] = A[i]; j++; } } for (int i = 0; i < j; i++) cout << A[i] << " "; } Above method deletes range of elements from an array. Please explain the logic of above code in simple english (algorithm and comments).
- Consider the following function: public void bSearch(int[] A, int value, int start, int end) { if (end value) { return bSearch(A, value, start, mid); } else if (value > A[mid]) { return bSearch(A, value, mid+1, end); } else { return mid; } } The mutation program P' changes the part as shown below. } else if (value > A[mid]) { return bSearch(A, value, mid+2, end); } else { return mid; } A test input is given t1: {A = [2, 3, 5, 7, 10, 11], value = 6, start=0, end=5}. Which statement below best describes the mutant with regards to the test input? O a. t1 resolves mutant P O b. The mutant P' with regards to test input t1 is live.// 9. Is It Prime? function isPrime(n) { if (n < 2 || n % 1 != 0) { return false; } for (let i if (n % i } return true; 2; i < n; ++i) { 0) return false; == }le.com/forms/d/e/1FAlpQLSc6PlhZGOLJ4LOHo5cCGEf9HDChfQ-tT1bES-BKgkKu44eEnw/formResponse The following iterative sequence is defined for the set of positive integers: Sn/2 3n +1 ifn is odd if n is even Un = Using the rule above and starting with 13, we generate the following sequence: 13 u13 = 40 u40 =20 u20 = 10→ u10 =5 u5 = 16 u16 = 8 ug = 4 → Us =2 u2 =1. It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. The below function takes as input an integer n and returns the number of terms generated by the sequence starting at n. function i-Seq (n) u=n; i=%3; while u =1 if statement 1 u=u/2; else statement 2 end i=i+1; end statement 1 and statement 2 should be replaced by: None of the choices statement 1 is "mod(u,2)=D%3D0" and statement 2 is "u = 3*u+1;" statement 1 is "u%2" and statement 2 is "u = 3*u+1;" O statement 1 is "mod(n,2)=30" and statement 2 is "u = 3*n+1;"
- // Assume all libraries are included void QQ(int n); // int main () { // Random questions QQ (5) ; 3 4 6. 7 8. 9. 10 return 0; 11 } // void QQ(int n) 12 13 14 { if(n >= 1) { 15 16 17 cout << n; QQ (--n); } else 18 19 20 21 cout << n; 22 } // 23int getMax(int arr[], int n) { intmx=arr[0]; for (inti=1; i<n; i++) if (arr[i] >mx) mx=arr[i]; returnmx; } Can u give me the code for this one as well....this is the first function and countsort is the second1. The runsing time of fl (n) and C) are
