How many times is the comparison i ≥ n performed in the following program?

int i=85, n=5;
main() {
while (i >= n) {







sumitverma's picture

The value of i − n is 80 initially. We run the loop as long as i − n ≥ 0 and in each iteration, i − n decreases by 2. Just before the kth time the comparison is performed (for k ≥ 1), the value of i−n is 80−2k+2. Hence just before the forty-first comparison, the value of i − n is 0. After the forty-first comparison, the loop is executed one last time. We need to make the comparison once more to exit the loop. Thus the correct answer is 42.

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