This content originally appeared on DEV Community and was authored by Rakesh Reddy Peddamallu
key - in order to start in circular way ->
generally we iterate starting from i=0 , j = (i+1) but since this is circular
we iterate i=0 , j = (i+1) % nums.length
in order to move to next element generally we do j++ or j = j +1
but since this is circular the next element of j will be (j+1) % nums.length
*Sample working *->
suppose when i goes to 3 , as j = i+1 ie 4 it crossed the end incase of nums = [1,2,1] , (i+1) % 4 ie 4 % 4 gives 0 , so it starts from index 0 instead of going to index 4
Learning
we can fill array of size n with -1 using new Array(n).fill(-1);
we can use break statements in for loops to exit out of loop
continue statement will skip one iteration
/**
* @param {number[]} nums
* @return {number[]}
*/
var nextGreaterElements = function(nums) {
const n = nums.length;
const result = new Array(n).fill(-1)
for (let i = 0; i < nums.length; i++) {
let j = (i + 1) % nums.length; // Start from next index in a circular manner
while (j !== i) {
if (nums[j] > nums[i]) {
result[i] = nums[j];
break; // Break out of the loop once we find the next greater element
}
j = (j + 1) % nums.length; // Move to the next index in a circular manner
}
}
return result;
};
Please do follow the series if you are struggling with leetcode questions 😇
This content originally appeared on DEV Community and was authored by Rakesh Reddy Peddamallu
Rakesh Reddy Peddamallu | Sciencx (2024-06-24T03:52:50+00:00) Leetcode – 503. Next Greater Element II. Retrieved from https://www.scien.cx/2024/06/24/leetcode-503-next-greater-element-ii/
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