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Compute the frequencycount of the following code fragment.for(i=0;i<n; i++)for(j=0;j<n; j++)printf(“%d”,a[i][j]);

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Solution

The frequency count of the given code fragment can be computed as follows:

  1. The outer loop runs 'n' times, where 'n' is the size of the array. This is because 'i' starts from 0 and increments by 1 until it reaches 'n'.

  2. Inside the outer loop, there is an inner loop that also runs 'n' times for the same reasons.

  3. Inside the inner loop, there is a printf statement that gets executed.

  4. Since the printf statement is inside the inner loop, which is inside the outer loop, the printf statement gets executed 'n' times for each iteration of the outer loop.

  5. Therefore, the total frequency count of the printf statement (i.e., the total number of times the printf statement gets executed) is 'n' (for the outer loop) times 'n' (for the inner loop), which equals 'n^2'.

So, the frequency count of the given code fragment is 'n^2'.

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