In binary search implementation, assigning mid to left (left = mid) instead of mid + 1 causes an infinite loop because the search space never shrinks; the correct fix is to use left = mid + 1 to ensure progress toward the target.
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Deep Dive
This Binary Search Code Runs Forever #binarysearch #dsa #coding #java #codinginterview #leetcodeAdded:
Look at this code. The binary search implementation can run forever. What I mean is it can get stuck in an infinite loop. At first glance, the code looks completely correct, right? I have taken a lot of interviews and I have seen many people make this exact mistake.
Let's focus specifically on this condition, right? Assigning mid value to the left and let me show you why this breaks. Let's assume the input array contains only three element elements and the target is eight. So, first we calculate the mid using the value of left and right variable and the mid is one. The value present at mid index is seven, which is smaller than eight. So, we move right. In second iteration, the value of left variable is one because as per the code, we are assigning the value of mid to the left variable. The value of right variable is two. Again, we recompute mid and its value is one. Wait, mid is still one?
Again, the value present at mid index is seven.
Again, left becomes one. Same state again. And after next iteration, again we are in a same state.
Now you are thinking in right direction.
We are stuck in an infinite loop. What's the fix? Instead of assigning mid, we can simply assign mid plus one to the left because we already checked mid and we know the value present at mid index is smaller than the target, right? This simple mistake leads to our code stuck in an infinite loop.
I explain all three binary search traps in detail in the full video. Check it out.
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