Review of week 280 of leetcode

6004. obtain 0 The number of operations

Here are two for you non-negative Integers num1 and num2 .

Each step operation   in , If num1 >= num2 , You have to use num1 reduce num2 ; otherwise , You have to use num2 reduce num1 .

for example ,num1 = 5 And num2 = 4 , Should use the  num1 reduce num2 , therefore , obtain num1 = 1 and num2 = 4 . However , If num1 = 4 And num2 = 5 , After one step operation , obtain num1 = 4 and num2 = 1 .
Return to make num1 = 0 or num2 = 0 Of Operands .

Simulation is enough

class Solution {
    public int countOperations(int num1, int num2) {
        if(num1==0||num2==0){
            return 0;
        }
        int ans=1;
        if(num1>=num2){
            num1-=num2;
        }else{
            num2-=num1;
        }
        ans+=countOperations(num1,num2);
        return ans;
    }
}

---

6005. The minimum number of operands to make an array an alternating array  

I'll give you a subscript from 0 Starting array nums , The array consists of n It's made up of four positive integers .

If the following conditions are met , The array nums It's a Alternating array ：

nums[i - 2] == nums[i] , among 2 <= i <= n - 1 .
nums[i - 1] != nums[i] , among 1 <= i <= n - 1 .
In one step operation in , You can choose the subscript i And will nums[i] change by any Positive integer .

Returns the that makes the array an alternating array The minimum number of operands .

Count the maximum number of occurrences corresponding to odd and even positions 、 Next largest number , If the maximum number is different , Use these numbers , The answer is len-max1-max2, If the same , Use the next largest occurrence number of another

class Solution {
    public int minimumOperations(int[] nums) {
        int n=nums.length,ans=0;
        if(n==1)return 0;
        int even1=0,even2=0,odd1=0,odd2=0;
        int max1=0,max2=0;
        Map<Integer,Integer>map1=new HashMap<>();
        Map<Integer,Integer>map2=new HashMap<>();
        for(int i=0;i<n;i+=2){
            map1.put(nums[i],map1.getOrDefault(nums[i],0)+1);
        }
        for(int num:map1.keySet()){
            int cnt=map1.get(num);
            if(cnt>even1){
                max1=num;
                even2=even1;
                even1=cnt;        
            }else if(cnt>even2){
                even2=cnt;
            }
        }
        for(int i=1;i<n;i+=2){
            map2.put(nums[i],map2.getOrDefault(nums[i],0)+1);
        }
        for(int num:map2.keySet()){
            int cnt=map2.get(num);
            if(cnt>odd1){
                max2=num;
                odd2=odd1;
                odd1=cnt;        
            }else if(cnt>odd2){
                odd2=cnt;
            }
        }
        if(max1!=max2){
            ans=n-even1-odd1;
        }else{
            if(odd1+even2>even1+odd2){
                System.out.println(" Use the maximum odd number "+odd1+" And sub large even numbers "+even2);
                ans=n-odd1-even2;
                
            }else{
                System.out.println(" Use the maximum even number "+even1+" And sub large odd numbers "+odd2);
                ans=n-even1-odd2;
            }
        }
        System.out.println(n);
        return ans;
    }
}

---

 6006. Take out the smallest number of magic beans

To give you one just   An array of integers  beans , Each integer represents the number of magic beans in a bag .

Please... From each bag   take out   Some beans （ It's fine too   Don't take out ）, Make the rest Non empty In the bag （ namely At least   also One   Magic bean bag ） Number of magic beans   equal  . Once the magic beans are removed from the bag , You can't put it in any other bag .

Please return to where you need to take out the magic beans Minimum number .

Prioritize , Then traverse from front to back , The calculation is based on the current bag ( Clear all the front ) The number of beans you need to take out ( Note that it is not possible to empty the last bag , The last bag as a benchmark must consume less beans than emptying ), Returns the minimum number of magic beans

class Solution {
    public long minimumRemoval(int[] beans) {
        Arrays.sort(beans);
        int n=beans.length;
        if(n==1)return 0;
        long sum=beans[0];
        long[]pre=new long[n];
        pre[0]=(long)beans[0];
        for(int i=1;i<n;i++){
            sum+=(long)beans[i];
        }
        for(int i=0;i<n;i++){
            pre[i]=sum-(long)(n-i)*beans[i];
        }
        Arrays.sort(pre);
        return pre[0];
    }
}

---