# Find next greater indexes
```java
public class Main {
public static void main(String[] args) {
int[] nums = new int[]{13, 8, 1, 5, 2, 5, 9, 7, 6, 12};
System.out.println(Arrays.toString(findNextGreaterIndexes(nums)));
}
private static int[] findNextGreaterIndexes(int[] nums) {
// Next greater indexes mean the values should be strictly increasing,
// so we use ">"
// Time Complexity: O(n), Iterating through all elements in array
// Space Complexity: O(n), In worst case, we need to store all elements in stack
// Edge case
if(nums==null) {
return null;
}
// Initialise an empty stack
Stack stack = new Stack<>();
// Create a result int[] where we will store next greatest indexes
int[] result = new int[nums.length];
Arrays.fill(result, -1); // Initialise all the values in result with -1
for(int i=0;i Numerical value of index on top of stack
// nums[i] --> current number
// Loop through stack and check
// Since need to find strictly increasing index, we check
// if the numeric value of index stored in top of stack < than current number
// This would help us find strictly increasing index
while(!stack.isEmpty() && nums[stack.peek()] < nums[i]) {
// As we found suitable value for the index on stack.peek(),
// We pop it, which means we found greater index value for it
int stackTopIdx = stack.pop();
// Store the current index at stacktopidx positiaaon in result
result[stackTopIdx] = i;
}
// Push the current index to stack
// By following above while loop, the monotonic property of stack is always maintained
stack.push(i);
}
return result;
}
}
```
### Understanding of \`while\` loop comparison
### Final Result
