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Maximum Gap in C++
Suppose we have an array, that is not sorted. We have to find the maximum difference between successive elements in its sorted form. We will return 0 if the array contains less than 2 elements. So if the array is like [12,3,9,1,17], then the output will be 6, as the sorted array will be [1,3,9,12,17] so 5 will be the maximum difference as difference between 3 and 9 is 6.
To solve this, we will follow these steps −
minVal := inf, maxCal := -inf
n := size of nums
if n < 2, then return 0;
for i in range 0 to n – 1 −
minVal := min of nums[i] and minVal
maxVal := max of nums[i] and maxVal
gap := celing of maxVal – minVal / n – 1
make one array called bucketMax of size n – 1, and fill this with –inf
make one array called bucketMin of size n – 1, and fill this with inf
for i in range 0 to n – 1 −
x := nums[i]
if x = minVal or x = maxVal, then skip next part, go for next iteration
idx := (nums[i] – minVal) / gap.
bucketMax[idx] := max of bucketMax[idx] and nums[i]
bucketMin[idx] := min of bucketMin[idx] and nums[i]
ret := 0
prev := minVal
for i in range 0 to n – 1
if bucketMax[i] = -inf and bucketMin[i] = inf, then skip next part, go for next iteration
ret := max of ret and bucketMin[i] – prev
prev := bucketMax[i]
return max of ret, maxVal - prev
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
typedef long long int lli;
class Solution {
public:
int maximumGap(vector<int>& nums) {
lli minVal = INT_MAX;
lli maxVal = INT_MIN;
int n = nums.size();
if(n < 2) return 0;
for(int i = 0; i < n; i++){
minVal = min((lli)nums[i], minVal);
maxVal = max((lli)nums[i], maxVal);
}
int gap = ceil((double)(maxVal - minVal) / (double)(n - 1));
vector <int> bucketMax(n - 1, INT_MIN);
vector <int> bucketMin(n - 1, INT_MAX);
for(int i = 0; i < n; i++){
int x = nums[i];
if(x == minVal || x == maxVal) continue;
int idx = (nums[i] - minVal) / gap;
bucketMax[idx] = max(bucketMax[idx], nums[i]);
bucketMin[idx] = min(bucketMin[idx], nums[i]);
}
lli ret = 0;
lli prev = minVal;
for(int i = 0; i < n - 1; i++){
if(bucketMax[i] == INT_MIN && bucketMin[i] == INT_MAX) continue;
ret = max(ret, bucketMin[i] - prev);
prev = bucketMax[i];
}
return max(ret, maxVal - prev);
}
};
main(){
Solution ob;
vector<int> v = {12,3,9,1,17};
cout << (ob.maximumGap(v));
}Input
[12,3,9,1,17]
Output
6
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