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Degree of an Array in C++
Suppose we have an array of non-negative integers called nums, the degree of this array is actually the maximum frequency of any one of its elements. We have to find the smallest possible length of a contiguous subarray of nums, that has the same degree as nums.
So, if the input is like [1,2,2,3,1], then the output will be 2, this is because the input array has a degree of 2 because both elements 1 and 2 appear twice. The subarrays that have the same degree − [1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2] The shortest length is 2. So answer will be 2.
To solve this, we will follow these steps −
- Define an array freq of size 50000 and fill this with 0
- max_ := 0
- for each n in nums
- (increase freq[n] by 1)
- max_ := maximum of max_ and freq[n]
- fill freq array with 0.
- min_ := size of nums
- for initialize i := 0, j := -1, size := size of nums, when j < size, do −
- if j >= 0 and freq[nums[j]] is same as max_, then −
- min_ := minimum of min_ and j - i + 1
- otherwise when j < size - 1, then −
- increase j by 1
- increase freq[nums[j]] by 1
- Otherwise
- Come out from the loop
- if j >= 0 and freq[nums[j]] is same as max_, then −
- return min_
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int findShortestSubArray(vector<int>& nums) {
vector<int> freq(50000, 0);
int max_ = 0;
for (const int n : nums)
max_ = max(max_, ++freq[n]);
fill(freq.begin(), freq.end(), 0);
int min_ = nums.size();
for (int i = 0, j = -1, size = nums.size(); j < size;) {
if (j >= 0 && freq[nums[j]] == max_)
min_ = min(min_, j - i + 1), --freq[nums[i++]];
else if (j < size - 1)
++freq[nums[++j]];
else
break;
}
return min_;
}
};
main(){
Solution ob;
vector<int> v = {1, 2, 2, 3, 1};
cout << (ob.findShortestSubArray(v));
}Input
{1, 2, 2, 3, 1}Output
2
Updated on: 04-Jul-2020
453 Views
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