Introduction to Algorithms
3rd Edition
ISBN: 9780262033848
Author: Thomas H. Cormen, Ronald L. Rivest, Charles E. Leiserson, Clifford Stein
Publisher: MIT Press
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Chapter 33.4, Problem 6E
Program Plan Intro
To modify the closest pair
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Given a sorted array nums, remove the duplicates in-place such that each element appears only once and returns the new length.
Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
Clarification:
Confused why the returned value is an integer but your answer is an array?
Note that the input array is passed in by reference, which means a modification to the input array will be known to the caller as well.
Internally you can think of this:
// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);
// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
print(nums[i]);
}
Example 1:
Input: nums = [1,1,2]
Output: 2, nums = [1,2]
Explanation: Your function should return length = 2, with the first two elements of nums being 1 and 2…
Rewrite this code, using modular programming style, and correcting any errors if necessary.
2. Optimize the algorithm by making the following modifications:
(a) After the first pass, the largest number is guaranteed to be in the highest- numbered element in the array; after the second pass, the two highest numbers are "in place," and so on. Instead of comparing every pair on every pass, modify the algorithm to make as few comparisons as necessary on each pass.
(b) Modify the algorithm to check at the end of each pass if any swaps have been made. If none have been made, the data must already be in the proper order, so the program should terminate. Observe the single-exit point rule when making this modification.
3. Insert appropriate statements in your code to output the state of the array after every pass.
#include <iostream>
#include <iomanip>
using namespace std;
using std::setw;
using std::cout;
using std::endl;
using std::size_t;
int main() {
const short…
.WAP to sort(arrange) the elements of array in increasing order.input: 22 5 6 88 9output: 5 6 9 22 88
Chapter 33 Solutions
Introduction to Algorithms
Ch. 33.1 - Prob. 1ECh. 33.1 - Prob. 2ECh. 33.1 - Prob. 3ECh. 33.1 - Prob. 4ECh. 33.1 - Prob. 5ECh. 33.1 - Prob. 6ECh. 33.1 - Prob. 7ECh. 33.1 - Prob. 8ECh. 33.2 - Prob. 1ECh. 33.2 - Prob. 2E
Ch. 33.2 - Prob. 3ECh. 33.2 - Prob. 4ECh. 33.2 - Prob. 5ECh. 33.2 - Prob. 6ECh. 33.2 - Prob. 7ECh. 33.2 - Prob. 8ECh. 33.2 - Prob. 9ECh. 33.3 - Prob. 1ECh. 33.3 - Prob. 2ECh. 33.3 - Prob. 3ECh. 33.3 - Prob. 4ECh. 33.3 - Prob. 5ECh. 33.3 - Prob. 6ECh. 33.4 - Prob. 1ECh. 33.4 - Prob. 2ECh. 33.4 - Prob. 3ECh. 33.4 - Prob. 4ECh. 33.4 - Prob. 5ECh. 33.4 - Prob. 6ECh. 33 - Prob. 1PCh. 33 - Prob. 2PCh. 33 - Prob. 3PCh. 33 - Prob. 4PCh. 33 - Prob. 5P
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- : In searching an element in an array, linear search can be used, even though simple to implement, but not efficient, with only O(n) time complexity. Assuming the array is already in sorted order, modify the search function below, using a better algorithm, so the average time complexity for the search function is O(log n). include <iostream> using namespace std; int search(int al), int s, int v) { 1/ Modify below codes. for (int i = 0; i <s; i++) { if (a[i] = v) return i; return -1; int main() { int intArray:10] = { 5, 7, 8, 9, 10, 12, 13, 15, 20, 34); // Search for element '12' in 10-elements integer array. cout << search(intArray, 10, 12); // '5' will be printed out. // Search for element '35' in 10-elements integer array. cout << search(intArray, 10, 35); // '-1' will be printed out. // Index '-l' means that the element is not found. return 0;arrow_forwardYou wish to find duplicates in an unsorted n-element array A. 1,..., 2n integers.arrow_forward01... ""Implementation of the Misra-Gries algorithm.Given a list of items and a value k, it returns the every item in the listthat appears at least n/k times, where n is the length of the array By default, k is set to 2, solving the majority problem. For the majority problem, this algorithm only guarantees that if there isan element that appears more than n/2 times, it will be outputed. If thereis no such element, any arbitrary element is returned by the algorithm.Therefore, we need to iterate through again at the end. But since we have filtredout the suspects, the memory complexity is significantly lower thanit would be to create counter for every element in the list. For example:Input misras_gries([1,4,4,4,5,4,4])Output {'4':5}Input misras_gries([0,0,0,1,1,1,1])Output {'1':4}Input misras_gries([0,0,0,0,1,1,1,2,2],3)Output {'0':4,'1':3}Input misras_gries([0,0,0,1,1,1]Output None"""..arrow_forward
- Pseudo Code shown in Figure Q1(a) is an algorithm for binary searching for an array with n number of elements. By applying this algorithm, show step by step approach on how to find number 11 in an array as depicted in Figure Q1(b). low + 0 high e n-1 while (low s high) do ix + (low + high) /2 if (t = Alix)) then return ix else if (t < A[ix]) then high e ix - 1 else low + ix + 1 return -1 Figure Ql(a) 2 4 6 9 11 12 | 25 [0] [1) [2] [3] (4) [5) [6] Figure Q1(b)arrow_forwardGiven an infix expression 2*3/(2-1)+5*3. Your task is to covert the givenexpression into postfix notation. The final result should be 23*21-/53*+. (Note: Make use ofstack data structure only. However the underlying implementation should be an array). Evaluatethe resultant postfix notation and show the final result. The final result should be 21arrow_forwardFollowing is the function for interpolation search. This searching algorithm estimates the position (index) of a key in array based on the elements in the first position and last position in the array, and the length of array. The array must be sorted in ascending order. Suppose array A contains the following 15 elements: A = [1, 3, 3, 10, 17, 22, 22, 22, 24, 25, 26, 27, 27, 28, 28] At first iteration, at which position (index) the element of 24 is estimated in array A? In which part of array (starting index and ending index) the searching should continue? How many iterations the searching are performed until the element of 24 is found? int InterpolationSearch(int x[], int key, int n) { int mid, min = 0, max = n-1; while(x[min] < key && x[max] > key) { mid = min + ((key-x[min])*(max-min)) / (x[max]-x[min]); if(x[mid] < key) min = mid + 1; else if(x[mid] > key) max = mid - 1; else return mid; } if…arrow_forward
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