To calculate the run-time efficiency of the given program segment, we need to analyze the number of iterations performed by each loop.

1. The outermost loop runs from i = 1 to n, incrementing i by 1 in each iteration. Therefore, it will execute n times.

2. The second loop runs from j = 1 to n, incrementing j by 1 in each iteration. Since this loop is nested inside the outer loop, it will execute n times for each iteration of the outer loop. So, the total number of iterations for this loop is n * n = n^2.

3. The innermost loop runs from k = 1 to n, incrementing k by 1 in each iteration. Since this loop is nested inside both the outer and second loops, it will execute n times for each iteration of the outer and second loops. So, the total number of iterations for this loop is n * n * n = n^3.

4. Inside the innermost loop, the print statement is executed. However, the time complexity of the print statement is considered constant, as it does not depend on the input size.

To calculate the overall run-time efficiency, we multiply the number of iterations of each loop. Therefore, the total number of iterations for the given program segment is n * n * n = n^3.

Hence, the run-time efficiency of the given program segment is O(n^3), where n represents the input size.

Question

To calculate the run-time efficiency of the given program segment, we need to analyze the number of iterations performed by each loop.

1. The outermost loop runs from i = 1 to n, incrementing i by 1 in each iteration. Therefore, it will execute n times.

2. The second loop runs from j = 1 to n, incrementing j by 1 in each iteration. Since this loop is nested inside the outer loop, it will execute n times for each iteration of the outer loop. So, the total number of iterations for this loop is n * n = n^2.

3. The innermost loop runs from k = 1 to n, incrementing k by 1 in each iteration. Since this loop is nested inside both the outer and second loops, it will execute n times for each iteration of the outer and second loops. So, the total number of iterations for this loop is n * n * n = n^3.

4. Inside the innermost loop, the print statement is executed. However, the time complexity of the print statement is considered constant, as it does not depend on the input size.

To calculate the overall run-time efficiency, we multiply the number of iterations of each loop. Therefore, the total number of iterations for the given program segment is n * n * n = n^3.

Hence, the run-time efficiency of the given program segment is O(n^3), where n represents the input size.

Knowee AI · Accepted Answer

To calculate the run-time efficiency of the given program segment, we need to analyze the number of iterations performed by each loop.

1. The outermost loop runs from i = 1 to n, incrementing i by 1 in each iteration. Therefore, it will execute n times.

2. The second loop runs from j = 1 to n, incrementing j by 1 in each iteration. Since this loop is nested inside the outer loop, it will execute n times for each iteration of the outer loop. So, the total number of iterations for this loop is n * n = n^2.

3. The innermost loop runs from k = 1 to n, incrementing k by 1 in each iteration. Since this loop is nested inside both the outer and second loops, it will execute n times for each iteration of the outer and second loops. So, the total number of iterations for this loop is n * n * n = n^3.

4. Inside the innermost loop, the print statement is executed. However, the time complexity of the print statement is considered constant, as it does not depend on the input size.

To calculate the overall run-time efficiency, we multiply the number of iterations of each loop. Therefore, the total number of iterations for the given program segment is n * n * n = n^3.

Hence, the run-time efficiency of the given program segment is O(n^3), where n represents the input size.

Calculate the run-time efﬁciency of the following program segment:for (i = 1; i <= n; i++)for (j = 1; j <= n; j++)for (k = 1; k <= n; k++)print ("%d %d %d\n", i, j, k);

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