702710. Maximum Gap
Maximum Gap
Slug: maximum-gap3845
Difficulty: Easy
Id: 702710
Topic Tags: Arrays, Data Structures
Company Tags: HunanAsset
Summary
Given an array of n integers, find the maximum gap between two consecutive numbers. The problem is about finding the largest difference between any pair of adjacent elements in the array.
Detailed Explanation
The approach to solve this problem involves iterating through the array and maintaining the minimum and maximum values seen so far. For each element, we update the minimum and maximum values if necessary. Then, we calculate the gap by subtracting the previous maximum value from the current minimum value. We keep track of the maximum gap found so far.
Here's a step-by-step breakdown of the solution:
- Initialize
maxSoFarto negative infinity andminSoFarto positive infinity. - Iterate through the array:
- For each element, update
minSoFarif it is smaller than the current minimum value. - Update
maxSoFarif it is larger than the current maximum value.
- For each element, update
- Calculate the gap by subtracting
maxSoFarfromminSoFar. - Keep track of the maximum gap found so far.
Time complexity: O(n), where n is the number of elements in the array. This is because we are iterating through the array once.
Space complexity: O(1), as we are only using a constant amount of space to store the minimum and maximum values.
Here's an ASCII art diagram illustrating the step-by-step process:
+---------------+
| arr |
+---------------+
|
|
v
+---------------+
| minSoFar | maxSoFar
+---------------+
|
|
v
Gap = maxSoFar - minSoFar
Update maxSoFar and minSoFar as needed
Optimized Solutions
Java
public int maximumGap(int[] arr) {
if (arr.length < 2) return 0;
int n = arr.length;
int[] minArr = new int[n];
int[] maxArr = new int[n];
for (int i = 0; i < n; i++) {
minArr[i] = arr[i];
maxArr[i] = arr[i];
}
for (int i = 1; i < n; i++) {
if (arr[i] < minArr[i-1]) minArr[i] = arr[i];
if (arr[i] > maxArr[i-1]) maxArr[i] = arr[i];
}
int maxGap = 0;
for (int i = 1; i < n; i++) {
maxGap = Math.max(maxGap, maxArr[i-1]-minArr[i]);
}
return maxGap;
}