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Python Articles
Page 99 of 855
Check if a triplet with given sum exists in BST in Python
Suppose, we are provided with a Binary Search Tree (BST) that contains integer values, and a number 'total'. We have to find out if there are any group of three elements in the provided BST where the addition of the three elements is equal to the supplied 'total' value.So, if the input is liketotal = 12, then the output will be True.To solve this, we will follow these steps −temp_list := a new list initialized with zeroinorder traverse the tree and put it in temp_listfor i in range 0 to (size of temp_list - 2), increase by 1, doleft := ...
Read MoreCheck if a word exists in a grid or not in Python
Suppose, we have a grid or a matrix of words. We have to check whether a given word is present in the grid or not. The word can be found in four ways, horizontally left and right and vertically up and down. If we can find the word we return True, otherwise False.So, if the input is likepghsfykdghtkghihnsjsojfghnrtyuinput_str = "python", then the output will be True.To solve this, we will follow these steps −Define a function find_grid() . This will take matrix, input_str, row_pos, col_pos, row_count, col_count, degreeif degree is same as size of input_str , thenreturn Trueif row_pos < ...
Read MoreCheck if frequency of all characters can become same by one removal in Python
Suppose we have a lowercase string s. We have to check whether the frequency of all characters are same after deleting one character or not.So, if the input is like s = "abbc", then the output will be True as we can delete one b to get string "abc" where frequency of each element is 1.To solve this, we will follow these steps −occurrence := a map with all characters of s and their frequenciesif occurrences of all characters in s are same, thenreturn Truefor each char in s, dooccurrence[char] := occurrence[char] - 1if occurrences of all characters in s ...
Read MoreCheck if frequency of character in one string is a factor or multiple of frequency of same character in other string in Python
Suppose we have two strings s and t, we have to check whether the occurrences of a character in s is multiple or a factor in t.So, if the input is like s = "xxyzzw" t = "yyyxxxxzz", then the output will be True as frequency of x in s is 2, and in t is 4, in s y is present only once, but in t there are three y's, there are same number of z in s and t and there is one w in s but not in t.To solve this, we will follow these steps −s_freq ...
Read MoreCheck if all levels of two trees are anagrams or not in Python
Suppose, we are provided with two binary trees. We have to check if each level of a binary tree is an anagram of the other binary tree's same level. We return True if it is an anagram, otherwise we return False.So, if the input is like, then the output will be True.To solve this, we will follow these steps −tree_1 is the root node of the first tree and tree_2 is the root node of the second tree.if tree_1 is same as null and tree_2 is same as null, thenreturn Trueif tree_1 is same as null or tree_2 is same ...
Read MoreCheck if frequency of each digit is less than the digit in Python
Suppose we have a number n, we have to check whether the occurrence of each digit of n is less than or equal to digit itself.So, if the input is like n = 5162569, then the output will be True as the digits and frequencies are (5, 2), (1, 1), (6, 2) and (9, 1), for all the frequency is either small or equal to the digit value.To solve this, we will follow these steps −for i in range 0 to 9, dotemp := n, cnt := 0while temp is non-zero, doif temp mod 10 is same as i, thencnt ...
Read MoreCheck if frequency of characters are in Recaman Series in Python
Suppose we have a lowercase string s. We have to check whether the occurrences of alphabets in s, after rearranging in any possible manner, generates the Recaman’s Sequence (ignoring the first term).The Recaman’s sequence is as follows −$$a_{n}=\begin{cases}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:0(if\:n=0) & \a_{n-1}-n(if\:a_{n}-n>0\wedge not\:present\in sequence) & \\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:a_{n-1}+n(otherwise)\end{cases}$$Some of the items of Recaman's Sequence are [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9, 24, ...] (The first term is ignored as this is 0)So, if the input is like s = "pppuvuuqquuu", then the output will be True as the characters and frequencies are (p, 3), (u, ...
Read MoreCheck if given array is almost sorted (elements are at-most one position away) in Python
Suppose we have an array of numbers called nums, where all elements are unique. We have to check whether nums is almost sorted or not. As we know an array is almost sorted when any of its elements can occur at a maximum of 1 distance away from its original position in the sorted array.So, if the input is like nums = [10, 30, 20, 40], then the output will be True as 10 is placed at its original place and all other elements are at max one place away from their actual position.To solve this, we will follow these ...
Read MoreCheck if given four integers (or sides) make rectangle in Python
Suppose we have a list of four sides, we have to check whether these four sides are forming a rectangle or not.So, if the input is like sides = [10, 30, 30, 10], then the output will be True as there are pair of sides 10 and 30.To solve this, we will follow these steps −if all values of sides are same, thenreturn Trueotherwise when sides[0] is same as sides[1] and sides[2] is same as sides[3], thenreturn Trueotherwise when sides[0] is same as sides[3] and sides[2] is same as sides[1], thenreturn Trueotherwise when sides[0] is same as sides[2] and sides[3] ...
Read MoreCheck if given number is a power of d where d is a power of 2 in Python
Suppose we a number n and another value x, we have to check whether it is a power of x or not, where x is a number power of 2.So, if the input is like n = 32768 x = 32, then the output will be True as n is x^3.To solve this, we will follow these steps −From the main method do the following −cnt := 0if n is not 0 and (n AND (n - 1)) is same as 0, thenwhile n > 1, don = n/2cnt := cnt + 1return cnt mod (log c base 2) is ...
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