Qualcomm interview question
Binary search
Interview Answer
Anonymous
14 Dec 2011
min := 1; max := N; {array size: var A : array [1..N] of integer} repeat mid := (min+max) div 2; if x > A[mid] then min := mid + 1; else max := mid - 1; until (A[mid] = x) or (min > max);
1