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Given an integer array arr, count element x such that x + 1 is also in arr. If there're duplicates in arr, count them seperately. https://leetcode.com/explore/challenge/card/30-day-leetcoding-challenge/528/week-1/3289/
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import java.util.*; | |
import java.util.stream.*; | |
class CountingConsecutiveElements { | |
public int countElements(int[] arr) { | |
Map<Integer, Integer> numberCounts = new HashMap<>(); | |
for(int num : arr) { | |
Integer wrappedNum = Integer.valueOf(num); | |
int currentCount = numberCounts.getOrDefault(wrappedNum,0); | |
numberCounts.put(wrappedNum,currentCount+1); | |
} | |
int result = 0; | |
for(Map.Entry<Integer, Integer> currentEntry: numberCounts.entrySet()) { | |
int currentNum = currentEntry.getKey().intValue(); | |
int currentCount = currentEntry.getValue().intValue(); | |
int relatedNumCount = numberCounts.getOrDefault(currentNum+1, 0); | |
// Even if there is only one x+1, count all x's. E.g. [1,2,2] | |
result += relatedNumCount == 0 ? 0 : currentCount; | |
} | |
return result; | |
} | |
} |
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