{"id":27283,"date":"2018-01-20T20:35:37","date_gmt":"2018-01-20T15:05:37","guid":{"rendered":"https:\/\/www.wikitechy.com\/technology\/?p=27283"},"modified":"2018-01-20T20:35:37","modified_gmt":"2018-01-20T15:05:37","slug":"python-algorithm-next-greater-element","status":"publish","type":"post","link":"https:\/\/www.wikitechy.com\/technology\/python-algorithm-next-greater-element\/","title":{"rendered":"Python Algorithm – Next Greater Element"},"content":{"rendered":"

Given an array, print the Next Greater Element (NGE) for every element. The Next greater Element for an element x is the first greater element on the right side of x in array. <\/span>Elements for which no greater element exist, consider next greater element as -1.<\/p>\n

Examples:
\na) <\/strong>For any array, rightmost element always has next greater element as -1.
\nb) <\/strong>For an array which is sorted in decreasing order, all elements have next greater element as -1.
\nc) <\/strong>For the input array [4, 5, 2, 25}, the next greater elements for each element are as follows.<\/p>\n

Element       NGE\r\n   4      -->   5\r\n   5      -->   25\r\n   2      -->   25\r\n   25     -->   -1\r\n<\/pre>\n
d)<\/strong> For the input array [13, 7, 6, 12}, the next greater elements for each element are as follows.<\/p>\n
  Element        NGE\r\n   13      -->    -1\r\n   7       -->     12\r\n   6       -->     12\r\n   12     -->     -1<\/pre>\n[ad type=”banner”]\n
Method 1 (Simple)<\/strong>
\nUse two loops: The outer loop picks all the elements one by one. The inner loop looks for the first greater element for the element picked by outer loop. If a greater element is found then that element is printed as next, otherwise -1 is printed.<\/p>\n
Python Programming:<\/strong><\/p>\n
 
 <\/div> 
# Function to print element and NGE pair for all elements of listdef printNGE(arr):     for i in range(0, len(arr), 1):         next = -1        for j in range(i+1, len(arr), 1):            if arr[i] < arr[j]:                next = arr[j]                break                     print(str(arr[i]) + " -- " + str(next)) # Driver program to test above functionarr = [11,13,21,3]printNGE(arr) # This code is contributed by Sunny Karira<\/code><\/pre> <\/div>\n
Output:<\/strong><\/p>\n
11 -- 13\r\n13 -- 21\r\n21 -- -1\r\n3 -- -1<\/pre>\n
Time Complexity: O(n^2). The worst case occurs when all elements are sorted in decreasing order.<\/p>\n[ad type=”banner”]\nMethod 2 (Using Stack)<\/strong>
\nThanks to pchild for suggesting following approach.
\n1) Push the first element to stack.
\n2) Pick rest of the elements one by one and follow following steps in loop.
\n\u2026.a) Mark the current element as next<\/em>.
\n\u2026.b) If stack is not empty, then pop an element from stack and compare it with next<\/em>.
\n\u2026.c) If next is greater than the popped element, then next <\/em>is the next greater element for the popped element.
\n\u2026.d) Keep popping from the stack while the popped element is smaller than next<\/em>. next<\/em> becomes the next greater element for all such popped elements
\n\u2026.g) If next <\/em>is smaller than the popped element, then push the popped element back.
\n3) After the loop in step 2 is over, pop all the elements from stack and print -1 as next element for them.<\/p>\n
Python Programming:<\/strong><\/p>\n
 
 <\/div> 
# Python program to print next greater element using stack # Stack Functions to be used by printNGE()def createStack():    stack = []    return stack def isEmpty(stack):    return len(stack) == 0 def push(stack, x):    stack.append(x) def pop(stack):    if isEmpty(stack):        print("Error : stack underflow")    else:        return stack.pop() '''prints element and NGE pair for all elements of   arr[] '''def printNGE(arr):    s = createStack()    element = 0    next = 0     # push the first element to stack    push(s, arr[0])     # iterate for rest of the elements    for i in range(1, len(arr), 1):        next = arr[i]         if isEmpty(s) == False:             # if stack is not empty, then pop an element from stack            element = pop(s)             '''If the popped element is smaller than next, then                a) print the pair                b) keep popping while elements are smaller and                   stack is not empty '''            while element < next :                print(str(element)+ " -- " + str(next))                if isEmpty(s) == True :                    break                element = pop(s)             '''If element is greater than next, then push               the element back '''            if  element > next:                push(s, element)         '''push next to stack so that we can find           next greater for it '''        push(s, next)     '''After iterating over the loop, the remaining       elements in stack do not have the next greater       element, so print -1 for them '''     while isEmpty(s) == False:            element = pop(s)            next = -1            print(str(element) + " -- " + str(next)) # Driver program to test above functionsarr = [11, 13, 21, 3]printNGE(arr) # This code is contributed by Sunny Karira<\/code><\/pre> <\/div>\n
Output:<\/strong><\/p>\n
 11 -- 13\r\n 13 -- 21\r\n 3 -- -1\r\n 21 -- -1\r\n<\/pre>\n
Time Complexity: O(n). The worst case occurs when all elements are sorted in decreasing order. If elements are sorted in decreasing order, then every element is processed at most 4 times.
\na) Initially pushed to the stack.
\nb) Popped from the stack when next element is being processed.
\nc) Pushed back to the stack because next element is smaller.
\nd) Popped from the stack in step 3 of algo.<\/p>\n[ad type=”banner”]\n","protected":false},"excerpt":{"rendered":"
Python Algorithm – Next Greater Element –  Stack – Given an array, print the Next Greater Element (NGE) for every element.<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,73012,4148,79607],"tags":[81326,81327,81325,81300,81328,81299,81329,81298,81302,81301],"class_list":["post-27283","post","type-post","status-publish","format-standard","hentry","category-coding","category-data-structures","category-python","category-stack","tag-next-greater-element-i-leetcode-java","tag-next-greater-element-ii-leetcode","tag-next-greater-element-iii","tag-next-greater-element-in-array-java","tag-next-greater-element-java","tag-next-greater-element-leetcode","tag-next-greater-element-leetcode-python","tag-next-greater-element-using-stack","tag-next-greater-element-using-stack-in-java","tag-next-greater-number"],"_links":{"self":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/posts\/27283","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/comments?post=27283"}],"version-history":[{"count":0,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/posts\/27283\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/media?parent=27283"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/categories?post=27283"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/tags?post=27283"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}