Problem
Given a non-empty array of non-negative integers nums, the degree of this array is defined as the maximum frequency of any one of its elements.
Your task is to find the smallest possible length of a (contiguous) subarray of nums, that has the same degree as nums.
Example 1:
Input: nums = [1,2,2,3,1]
Output: 2
Explanation:
The input array has a degree of 2 because both elements 1 and 2 appear twice.
Of 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 return 2.
Example 2:
Input: nums = [1,2,2,3,1,4,2]
Output: 6
Explanation:
The degree is 3 because the element 2 is repeated 3 times.
So [2,2,3,1,4,2] is the shortest subarray, therefore returning 6.
Constraints:
nums.lengthwill be between 1 and 50,000.nums[i]will be an integer between 0 and 49,999.
Solution
/**
* @param {number[]} nums
* @return {number}
*/
var findShortestSubArray = function(nums) {
var map = {};
var degree = 0;
for (var i = 0; i < nums.length; i++) {
if (!map[nums[i]]) {
map[nums[i]] = {
left: i,
right: i,
num: 1,
};
} else {
map[nums[i]].right = i;
map[nums[i]].num += 1;
}
degree = Math.max(degree, map[nums[i]].num);
}
var min = Number.MAX_SAFE_INTEGER;
for (var i = 0; i < nums.length; i++) {
if (map[nums[i]].num === degree) {
min = Math.min(map[nums[i]].right - map[nums[i]].left + 1, min);
}
}
return min;
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).