Maximum Gap
Problem Statement :
Given an integer array nums, return the maximum difference between two successive elements in its sorted form. If the array contains less than two elements, return 0. You must write an algorithm that runs in linear time and uses linear extra space. Example 1: Input: nums = [3,6,9,1] Output: 3 Explanation: The sorted form of the array is [1,3,6,9], either (3,6) or (6,9) has the maximum difference 3. Example 2: Input: nums = [10] Output: 0 Explanation: The array contains less than 2 elements, therefore return 0. Constraints: 1 <= nums.length <= 105 0 <= nums[i] <= 109
Solution :
Solution in C :
int cmpfunc (const void * a, const void * b) {
return ( *(int*)a - *(int*)b );
}
int maximumGap(int* nums, int numsSize){
qsort(nums, numsSize, sizeof(int), cmpfunc);
int max=0;
if(numsSize<2)
{
return 0;
}
for(int i=0;i<numsSize-1;i++)
{
if(abs(nums[i+1]-nums[i])>max)
{
max=abs(nums[i+1]-nums[i]);
}
}
return max;
}
Solution in C++ :
static auto speedup = [](){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
return 0;
}();
class Solution {
public:
int maximumGap(vector<int>& nums) {
int n=nums.size();
if(n<2)return 0;
int maxNum=INT_MIN;
int minNum=INT_MAX;
for(int i=0;i<n;i++)
{
maxNum=max(maxNum,nums[i]);
minNum=min(minNum,nums[i]);
}
if(maxNum==minNum){return 0;}
int gap=(maxNum-minNum)/(n-1);
if((maxNum-minNum)%(n-1) != 0){gap++;}
int minArr[n];
int maxArr[n];
for(int i=0;i<n;i++)
{
minArr[i]=INT_MAX;
maxArr[i]=INT_MIN;
}
for(int i=0;i<n;i++)
{
int bkt=(nums[i]-minNum)/gap;
minArr[bkt]=min(minArr[bkt],nums[i]);
maxArr[bkt]=max(maxArr[bkt],nums[i]);
}
int ans=INT_MIN;
int prev=INT_MIN;
for(int i=0;i<n;i++)
{
if(minArr[i]==INT_MAX){continue;}
if(prev==INT_MIN)
{
prev=maxArr[i];
}
else
{
ans=max(ans,minArr[i]-prev);
prev=maxArr[i];
}
}
return ans;
}
};
Solution in Java :
class Solution {
public int maximumGap(int[] nums) {
int n = nums.length;
if (n == 1) return 0;
int max = Integer.MIN_VALUE;
int min = Integer.MAX_VALUE;
for (int i = 0; i < nums.length; i++){
max = Math.max(max, nums[i]);
min = Math.min(min, nums[i]);
}
if (max == min) return 0;
int dx = (int)Math.ceil((double)(max-min)/(n-1));
int[] maxVal = new int[n-1];
int[] minVal = new int[n-1];
for (int i = 0; i < n-1; i++){
maxVal[i] = Integer.MIN_VALUE;
minVal[i] = Integer.MAX_VALUE;
}
for (int i = 0; i < n; i++){
if (nums[i]!= max && nums[i]!=min){
int index = (nums[i] - min)/dx;
maxVal[index] = Math.max(maxVal[index], nums[i]);
minVal[index] = Math.min(minVal[index], nums[i]);
}
}
int ans = Integer.MIN_VALUE;
int prev = min;
for (int i = 0; i < n-1; i++){
if (maxVal[i] != Integer.MIN_VALUE){
ans = Math.max(ans, minVal[i] - prev);
prev = maxVal[i];
}
}
ans = Math.max(ans, max - prev);
return ans;
}
}
Solution in Python :
class Solution:
def maximumGap(self, nums: List[int]) -> int:
h = list(set(nums))
h.sort()
if len(h)<2:return 0
m = 0
for i in range(len(h)-1):
if h[i+1]-h[i]>m:m=h[i+1]-h[i]
return m
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