Algorithm1(A[0 .... n– 1]) // Input: Array A[0 .... n– 1] of numbers count 0 for i= 0 to n do for j=i to n do if A[i] – A[j]] < 10 | count count + 1 return count
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- for (int i = 0;i>m2[i][j]; This C++ (2) ELä %3D :code used to Count from 0 to 2 O Print the array m2 O Read the array m2 O1. Estimate in terms of n, the worst case time complexity of each of the following pseudo-code snippets: a. ALGORITHM sum(n) sum -0 for i 0 to n do for j +0 to n*n do sum Esum + 1 return sumALGORITHM Unique Elements (A[0..n-1]) //Determines whether all the elements in a given array are distinct //Input: An array A[0..n-1] //Output: Returns "true" if all the elements in A are distinct and "false" otherwise // for i 0 to n - 2 do for ji+1 to n - 1 do if A[i] = A[j] return false return true A random list of 20 numbers is generated and used as input for the algorithm UniqueElements. Which of the following is NOT true when the algorithm terminates? OA. in the best case the first two elements are the same. B. in the worst case, there are no duplicate elements. C. in the worst case the number of comparisons is 190 D. in the worst case the number of comparisons is 400 Clear my choice
- ALGORITHM Unique Elements (A[0..n-1]) //Determines whether all the elements in a given array are distinct //Input: An array A[0..n - 1] //Output: Returns "true" if all the elements in A are distinct and "false" otherwise // for i 0 to n - 2 do for ji+1 to n - 1 do if A[i] = A[j] return false return true A random list of 20 numbers is generated and used as input for the algorithm Unique Elements. Which of the following is NOT true when the algorithm terminates? A. in the best case the first two elements are the same. B. in the worst case, there are no duplicate elements. C. in the worst case the number of comparisons is 190 D. in the worst case the number of comparisons is 400Python Numpy function to complete: def t19(N, s, X, y): Inputs: N: An integer s: A floating-point number - x: A floating-point number - y: A floating-point number Returns: A numpy array I of shape (N, N) such that I[i, j] exp(-||(j, i) –- (x, y)||^2 / s^2) Par: 3 lines Instructor: 2 lines return None// Input: array A[0, ..., n-1] minval :=A[0] maxval :=A[0] for i = 1 to n-1 do end for a. b. C. if A[i] < minval then minval :=A[i] end if if A[i] < maxval then maxval:=A[i] end if What does the algorithm do? What is the basic operations? What is the efficiency class of the algorithm?
- Complete the following function definition to recursively print the index of a unique value in an array or -1 if the value is not found:1 int getIndex(int *a, int s, int v) {2 if( ) { // if no values are found3 4 }5 if(a[s-1] == v) { // if a value is found6 7 }8 return getIndex( ); // recurse to check next value9 }Hint:Recurse through the array by counting down from size. 1. Complete the one line of code for line 2:2. Write one line of code for line 3:3. Write one line of code for line 6:4. Complete the one line of code for line 8: please send an atom code. not other softwareint g(int nums[], int n) { if (n == 1) return 0; int val = g(nums, n-1); if (nums[n-1] > nums[val]) return val; return n-1; } Give a recurrence T(n) for the number of times the code nums[n-1] > nums[val] is executed when the array values are in ascending order. Then solve the recurrenceOGiven an array arr[] of N non-negative integers representing the height of blocks. If width of each block is 1, compute how much water can be trapped between the blocks during the rainy season. Example 1: Input: N = 6 arr[] = {3,0,0,2,0,4} Output: 10.
- 9).An array of integers nums sorted in ascending order, find the startingand ending position of a given target value. If the target is not found in thearray, return [-1, -1]. For example:Input: nums = [5,7,7,8,8,8,10], target = 8Output: [3,5]Input: nums = [5,7,7,8,8,8,10], target = 11Output: [-1,-1]. Please weite your Code.* How many elements in the array A are * also in the array B? Assume B is sorted. 01: int overlap (int* A, int* B, int N) 02: { 03: int count = 0; 04: for (int i = 0; i < N; ++i) 05: { %3D 06: int x = A[i]; 07: int pos = lower_bound (B, B+N, x) - B: %3D if (pos <0 && B[pos] == x) 09: 10: 08: ++count; 11: 12: } 13: return count; 14:} According to the copy-and-paste technique, how would you annotate line 8 of this function by the time you had determined the complexity of lines 8-11?1.Given an array arr[] and an integer K where K is smaller than size of array, the task is to find the Kth smallest element in the given array. It is given that all array elements are distinct. Note :- l and r denotes the starting and ending index of the array. Example 1: Input: N = 6 arr[] = 7 10 4 3 20 15 K = 3 Output : 7.