2024 March 16 Problems

daryl...@gmail.com

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Mar 16, 2024, 12:40:48 PMMar 16

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Feel free to work on any of the problems you want; we'll have people present their solutions at 11:30.

Please post your solutions to this thread so others can use this as a reference.

2824Count Pairs Whose Sum is Less than Target87.0%Easy

2616Minimize the Maximum Difference of Pairs43.9%Medium

2790Maximum Number of Groups With Increasing Length19.4%Hard

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Flocela Maldonado

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Mar 16, 2024, 1:19:17 PMMar 16

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2824Count Pairs Whose Sum is Less than Target87.0%Easy

Time: O(n*lg(n)) Memory: O(1)

class Solution {
public:
    int countPairs(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end());

        int lt = 0;
        int rt = nums.size()-1;

        int sum = 0;
        while(lt < rt)
        {
            while(lt < rt && nums[lt] + nums[rt] >= target)
            {
              --rt;
            }
            sum += rt - lt;
            ++lt;
        }

        return sum;
    }
};

vinay

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Mar 16, 2024, 1:52:21 PMMar 16

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2616Minimize the Maximum Difference of Pairs43.9%Medium

Getting a memory limit exceeded owing to constraints, but here is one possible approach I could come up with:

time complexity o(nlogn) / dp could be NCp

space o(n*p)

class Solution {
    int[][]  memo = null;

    public int minimizeMax(int[] nums, int p) {
        Arrays.sort(nums);
        memo = new int[nums.length+1][p+1];
        for(int[] row : memo){
            Arrays.fill(row, -1);
        }
        return dp(0,p, nums);
    }

    public int dp(int i, int p, int[] nums){

        if(p == 0){
            return 0;
        }

        if(i >= nums.length-1){
            return (int)1e10;
        }

        if(memo[i][p] != -1){
            return memo[i][p];
        }

        int with = Math.max(Math.abs(nums[i]-nums[i+1]), dp(i+2, p-1,nums));
        int without =  dp(i+1, p, nums);

        return memo[i][p] = Math.min(with, without);
    }
}

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Priyank Gupta

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Mar 16, 2024, 4:21:20 PMMar 16

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# time : O(nlog(n))
# space: O(1)
class Solution:
    def countPairs(self, nums: List[int], target: int) -> int:
        nums = sorted(nums)
        start = 0
        end = len(nums)-1
        res = 0
        while(start < end):
            if nums[start] + nums[end] >= target:
                end -= 1
            else:
                res += (end-start)
                start += 1
        return res

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