Missing Number in a Sorted Array by hiteshmadapathi · Pull Request #1344 · super30admin/Competitive-Coding-1

hiteshmadapathi · 2026-04-15T23:59:32Z

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super30admin · 2026-04-16T00:06:47Z

Interview Problem: Find Missing Number in a sorted array (MissingNumbers.py)

Strengths:

You have identified that binary search is the right approach for this problem, which is good.
You have considered edge cases (first and last elements), which shows attention to boundary conditions.

Areas for Improvement:

The core logic of using binary search needs refinement. Instead of checking adjacent differences at every mid, you should use the fact that in a sorted array without duplicates, the difference between the value at an index and the index itself (starting from 1) should be consistent until the missing number. For example, in an array starting at 1, the value at index i should be i+1. So, if we compute (arr[mid] - mid), it should be 1 for all indices before the missing number and 2 after. This is the approach taken in the reference solution.
Your current code may cause index errors. For instance, when mid is 0, checking arr[mid-1] is invalid. Similarly, when mid is at the last index, checking arr[mid+1] is out of bounds. You need to ensure that your indices are within the array bounds.
The binary search loop condition and update steps are incorrect. The condition while low<=high might lead to an infinite loop because you set low=mid or high=mid without changing them by at least 1. You should update low to mid+1 or high to mid-1 to avoid infinite loops. The reference solution uses while (b - a) > 1 which ensures termination.
Consider the example: [1,2,4,5]. Your code might not correctly identify 3 as missing. When mid=1 (value=2), it checks adjacent: left is 1 (diff=1, ok), right is 4 (diff=2>1) so it returns 2+1=3. This works for this case. But for [1,3,4,5], when mid=1 (value=3), it checks left: 3-1=2>1 so returns 3-1=2 (correct). However, if the missing number is not adjacent to mid, your update step might not work. For example, [1,2,3,5,6,7] (missing 4). When mid=2 (value=3), adjacent differences are 1 (left) and 2 (right) so it would return 3+1=4? Actually, it checks arr[mid+1]-arr[mid] = 5-3=2>1, so returns 4. So it works for this. But the update step: you have if arr[high]-arr[mid] > high-mid: low=mid else high=mid. This might not be correct. For [1,2,3,5,6,7], initially low=0, high=5. Mid=2 (value=3). Then arr[high]-arr[mid]=7-3=4, high-mid=3, so 4>3 -> true, so set low=mid=2. Then next mid = (2+5)//2=3 (value=5). Then check adjacent: right: 6-5=1, left:5-3=2>1 -> returns 5-1=4? So it returns 4 correctly. But this update step is not standard and might fail for other cases.
However, your update step is not based on the value-index difference. The standard approach is to compare (arr[mid] - mid) with (arr[low] - low). If they are equal, the missing number is in the right half; else in the left half. This is more robust.
Also, your code does not handle the case when the array has only one element. For example, if arr=[2], n=2, then the missing number is 1. Your code checks if first element is not 1 -> returns 1. So it works. But if arr=[1], then n=2, so the missing number is 2. Your code checks last element !=2 -> returns 2. So edge cases are handled.
But the main issue is that your binary search might not terminate or might skip the missing number. For example, consider [1,2,3,4,6] (missing5). n=5. low=0, high=4. mid=2 (value=3). Adjacent differences are 1. Then check: arr[high]-arr[mid]=6-3=3, high-mid=2, so 3>2 -> true, so low=mid=2. Then next: low=2, high=4, mid=3 (value=4). Check adjacent: right:6-4=2>1 -> return 4+1=5. So it works. But if we have [1,3,4,5,6] (missing2). low=0, high=4. mid=2 (value=4). Check adjacent: left:4-3=1? Actually, arr[mid]-arr[mid-1]=4-3=1, so not >1. Right:5-4=1. Then update: arr[high]-arr[mid]=6-4=2, high-mid=2, so 2>2? false. So set high=mid=2. Then low=0, high=2. mid=1 (value=3). Check adjacent: left:3-1=2>1 -> return 3-1=2. So it works.
Despite these examples, the update step is not theoretically sound. It relies on the difference between the high value and mid value compared to the index difference. This might not work if the missing number is in the left half. For instance, consider [2,3,4,5] (missing1). n=5. Your code first checks arr[0]!=1 -> returns1. So it works. Similarly, [1,2,3,4] (missing5) -> returns5. So edge cases are handled.
But the update step: if arr[high]-arr[mid] > high-mid is equivalent to (arr[high]-arr[mid]) > (high-mid). This is true if there is a gap in the right half. But actually, if the entire array is continuous, this difference should be exactly (high-mid). So if it's greater, it means there is a gap in the right half. So it might work. However, it is not symmetric. For the left half, you don't have a similar check. In your code, you only update based on the right half condition. This might cause issues if the gap is in the left half.
Actually, in your code, you only have one condition for updating low and high. You always go to the mid if the right half has a gap? But what if the gap is in the left half? For example, [1,2,4,5,6] (missing3). low=0, high=4. mid=2 (value=4). Check adjacent: right:5-4=1, left:4-2=2>1 -> returns 4-1=3. So it returns correctly without using the update step. So the update step is not triggered in this case. So it works.
However, consider a larger array where the missing number is not adjacent to mid. For example, [1,2,3,4,5,6,8,9,10] (missing7). low=0, high=8. mid=4 (value=5). Adjacent differences are 1. Then update: arr[high]-arr[mid]=10-5=5, high-mid=4, so 5>4 -> true, so low=mid=4. Then low=4, high=8. mid=6 (value=8). Check adjacent: right:9-8=1, left:8-6=2>1? Actually, arr[mid-1] is 6? So 8-6=2>1 -> return 8-1=7. So it works.
But what if the missing number is at the beginning of the right half? It seems to work. However, the update step might cause an infinite loop if the array is large and the gap is exactly at

VERDICT: NEEDS_IMPROVEMENT

Interview Problem: Design Min Heap

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VERDICT: NEEDS_IMPROVEMENT

hiteshmadapathi added 5 commits December 31, 2021 02:51

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Missing Number in a Sorted Array#1344

Missing Number in a Sorted Array#1344
hiteshmadapathi wants to merge 5 commits intosuper30admin:masterfrom
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hiteshmadapathi commented Apr 15, 2026

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super30admin commented Apr 16, 2026

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hiteshmadapathi commented Apr 15, 2026

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super30admin commented Apr 16, 2026

Interview Problem: Find Missing Number in a sorted array (MissingNumbers.py)

Interview Problem: Design Min Heap

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