static int solution(int[] A) {
int N = A.length;
int result = 0;
Map<Integer, List<Integer>> map = new TreeMap<>();
for (int i = 0; i < N; i ) {
List<Integer> list = map.getOrDefault(A[i], new ArrayList<>());
list.add(i);
map.put(A[i], list);
}
for (Map.Entry<Integer, List<Integer>> entry : map.entrySet()) {
List<Integer> list = map.get(entry.getKey());
Collections.sort(list);
result = Math.max(result, (list.get(list.size() - 1) - list.get(0)));
}
return result;
}
With this above solution we can solve the problem but it is not O(N) time complexity. So I am looking for an optimized solution for this problem in Java.
CodePudding user response:
// Collections.sort(list);//Removing this line makes O(NlogK) to O(N) time complexity
static int solution(int[] A) {
int N = A.length;
int result = 0;
Map<Integer, List<Integer>> map = new HashMap<>();
for (int i = 0; i < N; i ) {
List<Integer> list = map.getOrDefault(A[i], new ArrayList<>());
list.add(i);
map.put(A[i], list);
}
for (Map.Entry<Integer, List<Integer>> entry : map.entrySet()) {
List<Integer> list = map.get(entry.getKey());
result = Math.max(result, (list.get(list.size() - 1) - list.get(0)));
}
return result;
}
CodePudding user response:
One of the solution is loop through all elements and keep track of only first and last occurence as follow:
static int solution(int arr[]) {
// Pair.first will contain the first occurrence and Pair.second will contain the last occurrence of number
HashMap<Integer, Pair<Integer, Integer>> minMaxMap = new HashMap<>();
for (int index = 0; index < arr.length; index ) {
Pair<Integer, Integer> existingValue = minMaxMap.get(arr[index]);
if (existingValue == null) {
minMaxMap.put(arr[index], new Pair<Integer, Integer>(index, index));
} else {
existingValue.second = index; // update the Pair.second to latest value.
}
}
int result = 0;
for (Pair<Integer, Integer> pair : minMaxMap.values()) {
int difference = pair.second - pair.first;
if (difference > result) {
result = difference;
}
}
return result;
}
In your solution, since we are using list, it requires more memory in case the array contains a lot of duplicate element. By avoiding list, you can even reduce the memory footprint.