runtime complexity
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I need help finding the runtime complexity Off the following code.
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- 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).a) FindMinIterative public int FindMin(int[] arr) { int x = arr[0]; for(int i = 1; i < arr.Length; i++) { if(arr[i]< x) x = arr[i]; } return x; } b) FindMinRecursive public int FindMin(int[] arr, int length) { if(length == 1) return arr[0]; return Math.Min(arr[length - 1], Find(arr, length - 1)); } What is the Big-O for this functions. Could you explain the recurisive more in details ?void radixSort(int arr[], int n) { intm=getMax(arr, n); for (intexp=1; m/exp>0; exp*=10) countSort(arr, n, exp); } void printData(int arr[], int start, int len) { if( start>=len ) return( 0 ); printf("%d\n", arr[start]); printData(arr, start+1, len); } int main() { intarr[]= {8, 9, 3, 4, 7, 2, 5, 6, 1}; intn=sizeof(arr) /sizeof(arr[0]); radixSort(arr, n); printData(arr, 0, n); return0; } In MIPS
- public class FindArrayDifference { static void arrayDifference(int a[], int b[]) { int k = 0; int [] c = new int[a.length]; for(int i=0; i < a.length; i++) { int j; for(j = 0; j < b.length; j++) if(a[i] == b[j]) break; if(j == b.length) c[k++] = a[i]; } for(int j = 0; j < k; j++) System.out.println(c[j]); } public static void main(String[] args) { int a[] = {1,2,3,7,8,15,26}; int b[] = {1,2,3,15,4,8,6}; arrayDifference(a,b); } } Calculate the algorithm step number and algorithm time complexity of the above program?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;}#include <stdio.h> int getMax(int arr[], int n) { intmx=arr[0]; for (inti=1; i<n; i++) if (arr[i] >mx) mx=arr[i]; returnmx; } void countSort(int arr[], int n, int exp) { intoutput[n]; inti, count[10] = { 0 }; for (i=0; i<n; i++) count[(arr[i] /exp) %10]++; for (i=1; i<10; i++) count[i] +=count[i-1]; for (i=n-1; i>=0; i--) { output[count[(arr[i] /exp) %10] -1] =arr[i]; count[(arr[i] /exp) %10]--; } for (i=0; i<n; i++) arr[i] =output[i]; } void radixSort(int arr[], int n) { intm=getMax(arr, n); for (intexp=1; m/exp>0; exp*=10) countSort(arr, n, exp); } void printData(int arr[], int start, int len) { if( start>=len ) return( 0 ); printf("%d\n", arr[start]); printData(arr, start+1, len); } int main() { intarr[]= {8, 9, 3, 4, 7, 2, 5, 6, 1}; intn=sizeof(arr) /sizeof(arr[0]); radixSort(arr, n); printData(arr, 0, n); return0; } Can I get the MIPS Assembly Language for the above C Program I have the MIPS Code for getMax and countSort functions So…
- using System;public static class Lab8{ public static void Main() { int[] compStats = new int[25]; int n = 0, large, small; double avg = 0; // Input values into the array InpArray(compStats, ref n); // Find the average of the elements in the array // *** Insert the code for the call to the FindAverage method // Find the largest element in the array // *** Insert the code for the call to the FindLarge method // Find the largest element in the array // *** Insert the code of the call to the FindSmall method // Print out the results Console.WriteLine("\nThe Average of the array is {0:F}", avg); // *** Insert the code to print out the largest and smallest values // Pause until user is done Console.ReadLine(); } // Method: InpArray // Description: Input values into an array. // Parameters: arrValues: the array to fill. // num: number of elements in the array. //…public class ReplaceEvenWithZero { public static void main(String[] args) { int[]a = new int[]{14, 2, 19, 3, 15, 22, 18, 7, 44, 39, 51, 78} ; // Print array a for (int i = 0; i < a.length; i++) { System.out.print(a[i] + " "); } System.out.println(); // Replace the even elements. replaceEven(a); // Print array a again to see new elements. for (int i = 0; i < a.length; i++) { System.out.print(a[i] + " "); } System.out.println(); } /** Replace the even elements in the given array with 0 @param arr the array to use for the replacements */ public static void replaceEven(int[] arr) { //-----------Start below here. To do: approximate lines of code = 2 // Write a for loop to go through each element, determine if it is even //If so, replace with 0. Hint: use the modulus operator % to determine if an…public class ArraySection { static void arraySection(int a[], int b[]) { int k = 0; int [] c = new int[a.length]; for(int i = 0; i < a.length; i++) { for(int j = 0; j < b.length; j++) if(a[i] == b[j]) c[k++] = a[i]; } for(int i = 0; i < k; i++) System.out.println(c[i]); System.out.println(); } public static void main(String[] args) { int a[] = { 1, 2, 3, 4, 5 }; int b[] = { 0, 2, 4,5 }; arraySection(a,b); }} Calculate the algorithm step number and algorithm time complexity of the above program?
- int binsearch (int X , int V [] , int n ) { int low , high , mid , i ; low = 0; high = n - 1; for ( i = 0; i < high ; i ++) { if( V[ i ] > V [ i +1]) return -2; } while ( low <= high ) { mid = ( low + high )/2; if ( X < V [ mid ]) high = mid - 1; else if ( X > V [ mid ]) low = mid + 1; else return mid ; } return -1; } This code takes as input a sorted array V of size n, and an integer X, if X exists in the array it will return the index of X, else it will return -1. 1. Draw a CFG for binsearch(). 2. From the CFG, identify a set of entry–exit paths to satisfy the complete statement coverage criterion. 3. Identify additional paths, if necessary, to satisfy the complete branch coverage criterion. 4. For each path identified above, derive their path predicate…#include int main() { int a[5]; for (int i = = a[i] 1; for (int j = 0; j = a) { } = int n *pa; while (n >= 1) { n = n / 2; printf("%d *pa); pa-- ; 2 1; } Q. What is the output?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: