To day let's see Leaders in array Problem

An element is leader if it is greater than all the elements to its right side.

we can solve this problem using

- two for-loop method(easy)
- for-loop method(efficient)

```
int[] findLeader(int[] arr) {
List<Integer> result = new ArrayList<>();
for (int i = 0; i < arr.length; i++) {
boolean isLeader = true;
for (int j = i + 1; j < arr.length; j++) {
if (arr[i] < arr[j]) {
isLeader = false;
break;
}
}
if (isLeader) {
result.add(arr[i]);
}
}
return result.stream().mapToInt(Integer::intValue).toArray();
}
```

In This method we are going to traverse the array using i loop

,we are going to check is there any element greater than the element present in the i index by using another for loop j

if we find any element greater than that we don't add that to list if not we add that element to the result

Time Complexity: O(n

^{2})

```
int[] findLeaderEfficient(int[] arr) {
List<Integer> result = new ArrayList<>();
int currentHigh = arr[arr.length - 1];
result.add(currentHigh);
for (int i = arr.length - 1; i > 0; i--) {
if (arr[i] > currentHigh) {
result.add(arr[i]);
currentHigh=arr[i];
}
}
return result.stream().mapToInt(Integer::intValue).toArray();
}
```

The element present at the last of the array will be in the result because we don't have any element at right side to that

we declare currentHigh with the last element value

we move from right to left

we check is the current number is greater than currentHigh

if yes we add that to result list and set that element as currentHigh, if not we move to the next element.

Time Complexity: O(n)

## Top comments (0)