Successful Pairs of Spells and Potions | LeetCode | Java

Approach 1 : Brute Force

This approach will give TLE(Time Limit Exceed). So, it forces us the optimize the solution.

class Solution {
    public int[] successfulPairs(int[] spells, int[] potions, long success) {

        int sLen = spells.length;
        int pLen = potions.length;

        int res[] = new int[sLen];

        for(int i=0; i<sLen; i++){
            int num = spells[i];
            int mul = 0;
            int count = 0;
            for(int j=0; j<pLen; j++){
                mul = num * potions[j];
                if(mul>=success){
                    count++;
                }
                mul = 0;
            }

            res[i] = count;
        }

        return res;
    }
}

Approach 2 : Binary Search

class Solution {
    public int[] successfulPairs(int[] spells, int[] potions, long success) {

        int sLen = spells.length;

        int pLen = potions.length;

        int res[] = new int[sLen];

        Arrays.sort(potions);

        for(int i=0; i<sLen; i++){
            int getCount = getCount(potions, spells[i], pLen, success);
            res[i] = getCount;
        }

        return res;
    }

    int getCount(int potions[], int num, int n, long success){
        int low = 0;
        int high = n-1;
        int mid = 0;

        while(low<=high){
             mid = low + (high-low)/2;

            long mul = (long)num * potions[mid];

            if(mul<success)
                low = mid+1;

            else
                high = mid-1;
        }

        return n-low;
    }
}

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