You can use a Stack and the time complexity is O(N).
The algorithm:
Start from the left side and move towards the right. When you pick an element from the array (let's say x), pop the Stack until the elements from the Stack (lets say y) has an element greater than the array element i.e. x> y. Then push the element i.e. x to stack.
e.g. {40,50,11,32,55,68,75} . here s is stack.
40, as s is empty push it. s: 40
50, as s.peek() < 50 so pop 40 (40's greater element is 50) then push 50. s: 50
Next higher element of 40 - 50.
11, s.peek() > 11 so push 11. s: 50, 11
32, s.peek() < 32, so pop the element and now it's 50 which is greater than 32 hence push 32. s: 50 ,32
Next higher element of 11 - 32.
55, s.peek() < 55, so pop the element i.e. 32 then pop next as well as 50 < 55, then push 55. s: 55.
Next higher element of 32 - 55.
Next higher element of 50 - 55.
68, s.peek() < 68 so pop it and push 68. s: 68
75, s.peek() < 75 so pop it and push 75 s:75.
Next higher element of 68 - 75.
As the array does not have any element no just pop the stack say that for all the elements inside the array does not have greater element i.e. -1.
Next higher element of 75 - -1.
The same algorithm in code:
public static void getNGE(int[] a) {
Stack s = new Stack();
s.push(a[0]);
for (int i = 1; i < a.length; i++) {
if (s.peek() != null) {
while (true) {
if (s.peek() == null || s.peek() > a[i]) {
break;
}
System.out.println(s.pop() + ":" + a[i]);
}
}
s.push(a[i]);
}
while (s.peek() != null) {
System.out.println(s.pop() + ":" + -1);
}
}
