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Question 1. What is the time complexity for the following code/program? 1.1 for (int i = 1; i <= n; i++){ for (int j = 1; j <= n; j++){ 1.2 int a = 0; for (int k = n; k >= 1; k--){ sum = i + j+k; cout << sum << endl;}}} for (i = 0; i < N; i++) { for (j=N; j>i; j--) { a = a +i+j;}} 1.3 for (int i = 1; i <= n; i++){ for (int j = 1; j <= 100; j++){ sum = i + j + k;}}

Question 1. What is the time complexity for the following code/program? 1.1 for (int i = 1; i <= n; i++){ for (int j = 1; j <= n; j++){ 1.2 int a = 0; for (int k = n; k >= 1; k--){ sum = i + j+k; cout << sum << endl;}}} for (i = 0; i < N; i++) { for (j=N; j>i; j--) { a = a +i+j;}} 1.3 for (int i = 1; i <= n; i++){ for (int j = 1; j <= 100; j++){ sum = i + j + k;}}

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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## Question 1: What is the time complexity for the following code/program? ### 1.1 ```cpp for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { for (int k = n; k >= 1; k--) { sum = i + j + k; cout << sum << endl; } } } ``` - **Explanation:** This code contains three nested loops, each dependent on `n`. The outer loop runs `n` times, the middle loop also runs `n` times for each iteration of the outer loop, and the innermost loop runs `n` times for each iteration of the middle loop. - **Time Complexity:** \(O(n^3)\) ### 1.2 ```cpp int a = 0; for (i = 0; i < N; i++) { for (j = N; j > i; j--) { a = a + i + j; } } ``` - **Explanation:** There are two nested loops. The outer loop runs `N` times. The inner loop runs `N - i` times for each iteration of the outer loop, decreasing as `i` increases. - **Time Complexity:** \(O(N^2)\) ### 1.3 ```cpp for (int i = 1; i <= n; i++) { for (int j = 1; j <= 100; j++) { sum = i + j; } } ``` - **Explanation:** The outer loop runs `n` times, and the inner loop always runs 100 times, regardless of `n`. - **Time Complexity:** \(O(n)\) ### 1.4 ```cpp for (int i = 0; i < n; i++) { for (int j = i; j > 0; j /= 2) { cout << j << endl; } } ``` - **Explanation:** The outer loop runs `n` times. The inner loop performs a logarithmic division, reducing `j` by half each time, leading to `O(log i)` iterations. - **Time Complexity:** \(O(n \log n)\) ### 1.5 **Description:** A binary search works by comparing the target value
Transcribed Image Text:## Question 1: What is the time complexity for the following code/program? ### 1.1 ```cpp for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { for (int k = n; k >= 1; k--) { sum = i + j + k; cout << sum << endl; } } } ``` - **Explanation:** This code contains three nested loops, each dependent on `n`. The outer loop runs `n` times, the middle loop also runs `n` times for each iteration of the outer loop, and the innermost loop runs `n` times for each iteration of the middle loop. - **Time Complexity:** \(O(n^3)\) ### 1.2 ```cpp int a = 0; for (i = 0; i < N; i++) { for (j = N; j > i; j--) { a = a + i + j; } } ``` - **Explanation:** There are two nested loops. The outer loop runs `N` times. The inner loop runs `N - i` times for each iteration of the outer loop, decreasing as `i` increases. - **Time Complexity:** \(O(N^2)\) ### 1.3 ```cpp for (int i = 1; i <= n; i++) { for (int j = 1; j <= 100; j++) { sum = i + j; } } ``` - **Explanation:** The outer loop runs `n` times, and the inner loop always runs 100 times, regardless of `n`. - **Time Complexity:** \(O(n)\) ### 1.4 ```cpp for (int i = 0; i < n; i++) { for (int j = i; j > 0; j /= 2) { cout << j << endl; } } ``` - **Explanation:** The outer loop runs `n` times. The inner loop performs a logarithmic division, reducing `j` by half each time, leading to `O(log i)` iterations. - **Time Complexity:** \(O(n \log n)\) ### 1.5 **Description:** A binary search works by comparing the target value
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