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Maximum Gap using Bucket Sort

Certification: Advanced Level Accuracy: 100% Submissions: 1 Points: 15

Write a Java program to find the maximum gap between two adjacent elements in an array after the array is sorted. If the array contains fewer than 2 elements, return 0. You must solve this in linear time and space complexity using the bucket sort technique.

Example 1
    
  
  • Input: nums = [3, 6, 9, 1]
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  • Output: 3
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  • Explanation: When sorted, the array becomes [1, 3, 6, 9]. The gaps between adjacent elements are: (3-1)=2, (6-3)=3, (9-6)=3. The maximum gap is 3.
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Example 2
    
  
  • Input: nums = [10, 7, 1, 20, 15]
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  • Output: 6
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  • Explanation: When sorted, the array becomes [1, 7, 10, 15, 20]. The gaps between adjacent elements are: (7-1)=6, (10-7)=3, (15-10)=5, (20-15)=5. The maximum gap is 6.
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Constraints
    
  
  • 0 <= nums.length <= 10^5
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  • 0 <= nums[i] <= 10^9
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  • Time Complexity: O(n)
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  • Space Complexity: O(n)
    •
AlgorithmsMicrosoftTCS (Tata Consultancy Services)
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Hint
Solution

Solution Hints

    
  
  • Use bucket sort approach to achieve linear time complexity
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  • Find the minimum and maximum values in the array
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  • Create n-1 buckets, where n is the array length
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  • Calculate the bucket size as (max-min)/(n-1)
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  • Place each number in its corresponding bucket
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  • For each bucket, keep track of the minimum and maximum values
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  • The maximum gap will be the maximum of (min value of current bucket - max value of previous non-empty bucket)
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Steps to solve by this approach:

 Step 1: Check if the array has fewer than 2 elements, return 0 if true.

 Step 2: Find the minimum and maximum values in the array.

 Step 3: Calculate the bucket size as (max-min)/(n-1) and determine the number of buckets needed.

 Step 4: Create buckets and initialize each with min value as MAX_VALUE and max value as MIN_VALUE.

 Step 5: Place each number in its corresponding bucket and update the min and max values for that bucket.

 Step 6: Iterate through non-empty buckets to find the maximum gap between the min value of current bucket and max value of previous non-empty bucket.

 Step 7: Return the maximum gap found.

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