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(Solved): Solve in C++ #Problem 1: Upper and Lower Bound Description: Given a sorted array A of size \( \ma ...



#Problem 1: Upper and Lower Bound
Description:
Given a sorted array A of size \( \mathrm{N} \) and a search value \( \mathrm{

Solve in C++

#Problem 1: Upper and Lower Bound Description: Given a sorted array A of size \( \mathrm{N} \) and a search value \( \mathrm{S} \), you have to find the upper and lower bound value for S. In the first line, you will be given \( \mathrm{N} \) and \( \mathrm{S} \). In the second line, you will be given \( \mathrm{N} \) values denoting the sorted array. You have to print the upper and lower bound respectively with a single blank space. Keep a new line character after each output. Print the answer using 0 based indexing. Upper Bound: The index of the smallest value in the sorted array which is greater than S. For repeating such values consider the largest or the right-most index. If the largest value in the array is smaller than \( \mathrm{S} \). Then the upper bound is the sizo of the array. Lower Bound: The index of the largest value in the sorted array which is smaller or equal to S. For repeating such values consider the smallest or the left-most index. If \( \mathrm{S} \) is smaller than the smallest value In the array, consider the lower bound as 0 . Limits: \[ 1<=|\mathrm{A}|<=100000 \] Test Cases: - First solve for non repeating elements - Then consider the solution for repeating elements - Try to write code using separate functions


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CODE : #include using namespace std; int lower_bound(int a[], int s, int n) { if (s < a[0]) return 0; int idx, l = 0, r = n - 1; while
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