Maximum Gap Problem Statement

Given an array of n non-negative integers, find the maximum difference between 2 consecutive numbers in the sorted array of the given integers.

Example One

{
"arr": [1, 3, 7, 2]
}

Output:

4

We can see that in sorted order, the array becomes [1, 2, 3, 7], and the maximum difference is 7 - 3 = 4.

Example Two

{
"arr": [1, 1, 1]
}

Output:

0

Here the sorted array is [1, 1, 1], and so the maximum difference is 1 - 1 = 0.

Notes

• We need a linear time solution, not n * log(n). You can use up-to linear extra space.
• Assume input is always valid, and won't overflow.
• If the input has only one element, then return 0.
• There can be duplicates elements.

Constraints:

• 1 <= n <= 10⁶
• 0 <= array element <= 10⁶

We have provided one solution for this problem.

Maximum Gap Solution: Optimal

• First we find the minimum and maximum numbers; let's name them, min and max respectively.
• Now the range of the numbers is max – min + 1.
• We create an array named buckets of this size and use it to keep frequency of occurrence of numbers in an array.
• When we encounter the number present in the array, we increment buckets[arr[i] – max] by 1.
• After all the numbers in the array are processed, we find the maximum number of consecutive 0's between 2 non-zero elements of buckets array. That is the answer.

Time Complexity

O(n + range).

Auxiliary Space Used

O(range).

Space Complexity

O(n + range).

Input takes O(n) space, and we use O(range) of auxiliary space.

Code For Maximum Gap Solution: Optimal

    /*
    Asymptotic complexity in terms of \`n\` = length of the given array \`arr\` and
    \`range\` = max(Elements of array) – min(Elements of array):
    * Time: O(n + range).
    * Auxiliary space: O(range).
    * Total space: O(n + range).
    */

    static Integer maximum_gap(ArrayList<Integer> arr) {
        int n = arr.size();
        if(n==1) {
            return 0;
        }

        int max = 0;
        int min = 1000001;
        // Find maximum and minimum elements.
        for(int i = 0; i<n; i++) {
            min = Math.min(min,arr.get(i));
            max = Math.max(max,arr.get(i));
        }

        // Create array of size range to store the frequencies of the array elements.
        int range = max - min + 1;
        int buckets[] = new int[range + 5];
        for(int i : arr) {
            buckets[i - min]++;
        }

        // Finding maximum number of consecutive zeros between 2 non zero elements of buckets array.
        int ans = 0;
        for(int i = range-1; i>0;) {
            int cnt = 0, temp = i;
            i--;
            while(i>=0 && buckets[i]==0) {
                cnt++;
                i--;
            }
            if(i>=0) {
                ans = Math.max(ans, temp - i);
            }
        }
        return ans;
    }

We hope that these solutions to maximum gap have helped you level up your coding skills. You can expect problems like these at top tech companies like Amazon and Google.

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