2170. Minimum Operations to Make the Array Alternating LeetCode Solution
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Problem Statement of Minimum Operations to Make the Array Alternating
You are given a 0-indexed array nums consisting of n positive integers.
The array nums is called alternating if:
nums[i – 2] == nums[i], where 2 <= i <= n – 1.
nums[i – 1] != nums[i], where 1 <= i <= n – 1.
In one operation, you can choose an index i and change nums[i] into any positive integer.
Return the minimum number of operations required to make the array alternating.
Example 1:
Input: nums = [3,1,3,2,4,3]
Output: 3
Explanation:
One way to make the array alternating is by converting it to [3,1,3,1,3,1].
The number of operations required in this case is 3.
It can be proven that it is not possible to make the array alternating in less than 3 operations.
Example 2:
Input: nums = [1,2,2,2,2]
Output: 2
Explanation:
One way to make the array alternating is by converting it to [1,2,1,2,1].
The number of operations required in this case is 2.
Note that the array cannot be converted to [2,2,2,2,2] because in this case nums[0] == nums[1] which violates the conditions of an alternating array.
2170. Minimum Operations to Make the Array Alternating LeetCode Solution in C++
struct T {
unordered_map<int, int> count;
int mx = 0;
int secondMax = 0;
int maxFreq = 0;
int secondMaxFreq = 0;
};
class Solution {
public:
int minimumOperations(vector<int>& nums) {
// 0 := odd indices, 1 := even indices
vector<T> ts(2);
for (int i = 0; i < nums.size(); ++i) {
T& t = ts[i % 2];
const int freq = ++t.count[nums[i]];
if (freq > t.maxFreq) {
t.maxFreq = freq;
t.mx = nums[i];
} else if (freq > t.secondMaxFreq) {
t.secondMaxFreq = freq;
t.secondMax = nums[i];
}
}
if (ts[0].mx == ts[1].mx)
return nums.size() - max(ts[0].maxFreq + ts[1].secondMaxFreq,
ts[1].maxFreq + ts[0].secondMaxFreq);
return nums.size() - (ts[0].maxFreq + ts[1].maxFreq);
}
}; /* code provided by PROGIEZ */
2170. Minimum Operations to Make the Array Alternating LeetCode Solution in Java
class T {
public Map<Integer, Integer> count = new HashMap<>();
public int mx = 0;
public int secondMax = 0;
public int maxFreq = 0;
public int secondMaxFreq = 0;
}
class Solution {
public int minimumOperations(int[] nums) {
T[] ts = new T[2];
ts[0] = new T();
ts[1] = new T();
for (int i = 0; i < nums.length; ++i) {
T t = ts[i % 2];
t.count.merge(nums[i], 1, Integer::sum);
final int freq = t.count.get(nums[i]);
if (freq > t.maxFreq) {
t.maxFreq = freq;
t.mx = nums[i];
} else if (freq > t.secondMaxFreq) {
t.secondMaxFreq = freq;
t.secondMax = nums[i];
}
}
if (ts[0].mx == ts[1].mx)
return nums.length -
Math.max(ts[0].maxFreq + ts[1].secondMaxFreq, ts[1].maxFreq + ts[0].secondMaxFreq);
return nums.length - (ts[0].maxFreq + ts[1].maxFreq);
}
} // code provided by PROGIEZ
2170. Minimum Operations to Make the Array Alternating LeetCode Solution in Python
class T:
def __init__(self):
self.count = collections.Counter()
self.mx = 0
self.secondMax = 0
self.maxFreq = 0
self.secondMaxFreq = 0
class Solution:
def minimumOperations(self, nums: list[int]) -> int:
# 0 := odd indices, 1 := even indices
ts = [T() for _ in range(2)]
for i, num in enumerate(nums):
t = ts[i % 2]
t.count[num] += 1
freq = t.count[num]
if freq > t.maxFreq:
t.maxFreq = freq
t.mx = num
elif freq > t.secondMaxFreq:
t.secondMaxFreq = freq
t.secondMax = num
if ts[0].mx == ts[1].mx:
return len(nums) - max(ts[0].maxFreq + ts[1].secondMaxFreq,
ts[1].maxFreq + ts[0].secondMaxFreq)
return len(nums) - (ts[0].maxFreq + ts[1].maxFreq) # code by PROGIEZ
