We are given a circular array, print the next greater number for every element. If it is not found print -1 for that number. To find the next greater number for element Ai , start from index i + 1 and go uptil the last index after which we start looking for the greater number from the starting index of the array since array is circular
Give some hint for stack deque challenge 1st task hint statement is given in the description
hello @shrutikatyal
This approach makes use of a stack. This stack stores the indices of the appropriate elements from nums array. The top of the stack refers to the index of the Next Greater Element found so far. We store the indices instead of the elements since there could be duplicates in the nums array. The description of the method will make the above statement clearer.
We start traversing the numsnums array from right towards the left. For an element nums[i] encountered, we pop all the elements stack[top] from the stack such that nums[stack[top]] ≤ nums[i]. We continue the popping till we encounter a stack[top] satisfying nums[stack[top]]>nums[i]. Now, it is obvious that the current stack[top] only can act as the Next Greater Element for nums[i](right now, considering only the elements lying to the right of nums[i]).
If no element remains on the top of the stack, it means no larger element than nums[i] exists to its right. Along with this, we also push the index of the element just encountered(nums[i]), i.e. ii over the top of the stack, so thatnums[i](or stack[topstack[top) now acts as the Next Greater Element for the elements lying to its left.
We go through two such passes over the complete nums array. This is done so as to complete a circular traversal over the nums array. The first pass could make some wrong entries in the res array since it considers only the elements lying to the right of nums[i], without a circular traversal. But, these entries are corrected in the second pass