Determine the code below, tell me exactly how many times is
the operation sum++ performed ?
for ( i = 0; i < 100; i++ )
for ( j = 100; j > 100 - i; j--)
sum++;
Answer Posted / abdur rab
for ( i = 0; i < 100; i++ )
for ( j = 100; j > 100 - i; j--)
sum++;
first iteration i = 0
j = 100
j loop is executed untill ( 100 - i ) (ie 100 - 0 = 100 )
so output is 0 ( sum is incremented 0 times )
second iteration i = 1
j = 100
j loop is executed untill ( 100 - i ) (ie 100 - 1 = 99 )
so output is 1 ( sum is incremented 1 times )
third iteration i = 2
j = 100
j loop is executed untill ( 100 - i ) (ie 100 - 2 = 98 )
so output is 1 ( sum is incremented 2 times )
0 + 1 + 2 + 3.......+ 99 = ( n (n+1) ) / 2
( 99 (99+1) ) / 2 = 4950
| Is This Answer Correct ? | 15 Yes | 1 No |
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