1512. Number of Good Pairs

Problem Statement

Pattern:

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Solution

public int numIdenticalPairs (int[] nums){
	HashMap<Integer, Integer> map = new HashMap<>();
	// count how many times has this element been repeated in the past for each repetition.
	int totalPrevDuplicates=0;
	for (int num : nums) {
		int freq = map.getOrDefault(num, 0);
		// no. of times num[i] has appeared before added to total
		totalPrevDuplicates += freq;
		map.put(num, ++freq);
	}
	return totalPrevDuplicates;
}

Notes

• Its just like the No. of Handshakes Question
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• Here instead of no. of handshakes you have to count, how many pairs does each element form with thos in front of it / how many times is an element repeated in front of it
• Or look at it in reverse

Concept

How many elements before this can pair with this element / How many elements before this are duplicates of this element