Suppose we have postfix expression and we have to evaluate the value. Postfix expression is also known as Reverse polish notation. Here we have to use the stack data structure to solve the postfix expressions.So if the expression is “21+3*”, then the answer will be 9.Let us see the steps −for each character ch in the postfix expression, doif ch is an operator $\odot$ , thena := pop first element from stack, b := pop second element from the stackres := b $\odot$ apush res into the stackelse if ch is an operand, thenadd ch into the stackreturn element of ... Read More

Suppose we have two lists of same length they are called weights and values, and we also have another value capacity. Here weights[i] and values[i] represent the weight and value of the ith item. If we can take at most capacity weights, and that we can take any number of copies for each item, we have to find the maximum amount of value we can get.So, if the input is like weights = [1, 2, 3], values = [1, 5, 3], capacity = 5, then the output will be 11To solve this, we will follow these steps −Define a function ... Read More

Suppose we have a positive number n, we have to find the least number of perfect square numbers whose sum is same as n. So if the number is 10, then the output is 2, as the numbers are 10 = 9 + 1.To solve this, we will follow these steps −create one table for dynamic programming, of length n + 1, and fill it with infinitydp[0] := 0for i := 1, when i*i dp(n+1,INF);       dp[0] = 0;       for(int i =1;i*i

Suppose we have two lists of positive numbers called coins and salaries. Here coins[i] indicates the value for coin i and salaries[j] indicates the least amount of salary required to pay for worker j. Now suppose we have one coin per type and we must give each worker exactly one coin, we have to compute the number of ways to give coins to each worker. Here two ways are different if some worker receives one type of coin in one way but a different type of coin in the other way. If the result is very large return result mod ... Read More

Suppose we have a binary tree containing values 0, 1 and 2. The root has at least one 0 node and one 1 node. Now suppose there is an operation where we delete an edge in the tree and the tree becomes two different trees. We have to find the number of ways we can delete one edge such that none of the two trees contain both a 0 node and a 1 node.So, if the input is likethen the output will be 1 as we can only delete the 0 to 2 edge.To solve this, we will follow these ... Read More

Suppose we have a binary tree, we have to find a list of two numbers where the first number is the count of leaves in the tree and the second number is the count of non-leaf nodes.So, if the input is likethen the output will be (3, 2), as there are 3 leaves and 2 non-leaf nodes.To solve this, we will follow these steps −if n is null, thenreturn (0, 0)if left of n is null and right of n is null, thenreturn (1, 0)left := solve(left of n)right := solve(right of n)return (left[0] + right[0], 1 + left[1] + ... Read More

Suppose we have a binary tree where each node contains a digit from 0-9, we have to check whether its in-order traversal is palindrome or not.So, if the input is likethen the output will be True, as its inorder traversal is [2, 6, 10, 6, 2].To solve this, we will follow these steps −if root is null, thenreturn Truestack := a new stackcurr := rootinorder := a new listwhile stack is not empty or curr is not null, dowhile curr is not null, dopush curr into stackcurr := left of currnode := popped element from stackinsert value of node at ... Read More

Suppose we have a string s, we have to check whether any permutation of s is a palindrome or not.So, if the input is like s = "admma", then the output will be True, as we can rearrange "admma" to "madam" which is a palindrome.To solve this, we will follow these steps −c := a map holding each individual character count of scount := 0for each i in list of all values of c, doif i is odd, thenif count is same as 0, thencount := count + 1come out from the loopreturn Falsereturn TrueLet us see the following implementation ... Read More

Suppose we have a linked list. We have to check whether the list elements are forming a palindrome or not. So if the list element is like [5, 4, 3, 4, 5], then this is a palindrome, but a list like [5, 4, 3, 2, 1] is not a palindrome.To solve this, we will follow these steps −fast := head, slow := head, rev := None and flag := 1if head is empty, then return truewhile fast and next of fast is availableif next of the next of fast is available, then set flag := 0 and break the loopfast ... Read More

Suppose we have a linked list. We have to swap every two adjacent nodes (pair) and return its head. Here the constraint is that, we cannot modify the value of the nodes, only the node itself can be changed. So if the list is like [1, 2, 3, 4], then the resultant list will be [2, 1, 4, 3].To solve this, we will follow these steps −if head is not present, then return headfirst := head, second := next of head, dummy is one new node with value -1next of dummy := first, and prev := dummywhile second is not ... Read More