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Head of Career Skills Development & Coaching

*Based on past data of successful IK students

Given an array of `n`

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

```
{
"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.

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

Output:

```
0
```

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

- 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`

<= 10^{6} - 0 <= array element <= 10
^{6}

We have provided one solution for this problem.

- 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.

O(n + range).

O(range).

O(n + range).

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

```
/*
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.

If you are preparing for a tech interview at FAANG or any other Tier-1 tech company, register for Interview Kickstart's FREE webinar to understand the best way to prepare.

Interview Kickstart offers interview preparation courses taught by FAANG+ tech leads and seasoned hiring managers. Our programs include a comprehensive curriculum, unmatched teaching methods, and career coaching to help you nail your next tech interview.

We offer 18 interview preparation courses, each tailored to a specific engineering domain or role, including the most in-demand and highest-paying domains and roles, such as:

- Back-end Engineering Interview Course
- Front-end Engineering Interview Course
- Full Stack Developer Interview Course

To learn more, register for the FREE webinar.

Note: Input and Output will already be taken care of.

Given an array of `n`

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

```
{
"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.

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

Output:

```
0
```

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

- 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`

<= 10^{6} - 0 <= array element <= 10
^{6}

We have provided one solution for this problem.

- 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.

O(n + range).

O(range).

O(n + range).

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

```
/*
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.

If you are preparing for a tech interview at FAANG or any other Tier-1 tech company, register for Interview Kickstart's FREE webinar to understand the best way to prepare.

Interview Kickstart offers interview preparation courses taught by FAANG+ tech leads and seasoned hiring managers. Our programs include a comprehensive curriculum, unmatched teaching methods, and career coaching to help you nail your next tech interview.

We offer 18 interview preparation courses, each tailored to a specific engineering domain or role, including the most in-demand and highest-paying domains and roles, such as:

- Back-end Engineering Interview Course
- Front-end Engineering Interview Course
- Full Stack Developer Interview Course

To learn more, register for the FREE webinar.

**Attend our free webinar to amp up your career and get the salary you deserve.**

Hosted By

Ryan Valles

Founder, Interview Kickstart

- Designed by 500 FAANG+ experts
- Live training and mock interviews
- 17000+ tech professionals trained

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