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# EFFICIENCY QUESTIONS AND ANSWERS - EFFICIENCY QUESTIONS AND...

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EFFICIENCY QUESTIONS AND ANSWERS What is the Big-oh of the following code snippets? 1. for(int i = 0; i < n; i++) for(int j = 0; j < n; j++) 1+1; Answer: O(n^2) Explanation: The outer for loop has i = 0…n making it run n. The inner for loop has j = 0…n making it run n as well. Since the for loop with j is nested, we multiply their run times to get n^2. 2. for(int i = 0; i < n; i++) 1+1; for(int j = 0; j < n*n; j++) 1+1; Answer: O(n^2) Explanation: The loop with i runs from 0…n, making its runtime n. The loop with j runs from 0…n^2, making its runtime n^2. However, since the for loop with j is a separate loop, we merely add them together: n+n^2. We drop the lower terms to get O(n^2). 3. for(int i = 0; i < n*n; i++) for(int j = 1; j <= i*n; j++) for(int k = 0; k < j*j; k++) for(int r = 0; r < i; r++) 1+1; Answer: n^13 Explanation: The loop with i runs from 0…n*n, making it have a runtime of n^2. For the loop with j, the runtime is i*n, but i is dependent on n (i = n^2). Therefore, the loop with j has the runtime of n^3. The loop with k has runtime j^2, but j = n^3, so the runtime of this loop is (n^3)^2 = n^6. The

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EFFICIENCY QUESTIONS AND ANSWERS - EFFICIENCY QUESTIONS AND...

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