Server Side Programming Articles - Page 1498 of 2646

Suppose we have two values start and end, we have to find the minimum number of operations needed to convert start to end by using these operations −Decrement by 1Multiply by 2So, if the input is like start = 2, end = 7, then the output will be 3, as we can multiply 2 to get 4, then multiply 2 to get 8 and then subtract 1 to get 7.To solve this, we will follow these steps −ans := 0Do the following infinitely, doif end

Suppose we have one binary tree. We have to check whether the tree is symmetric tree or not. A tree will be said to be symmetric if it is same when we take the mirror image of it. From these two trees, the first one is symmetric, but second one is not.To solve this, we will follow these steps.We will call following steps recursively. The function will be solve(root, root)if the node1 and node2 are empty, then return trueif either node1 or node2 is empty, then return falsereturn true when node1.val = node2.val and solve(node1.left, node2.right) and solve(node1.right, node2.left)Let us ... Read More

Suppose we have two trees, we have to check whether we can transform first tree into second one by swapping any node's left and right subtrees any number of times.So, if the input is likethen the output will be TrueTo solve this, we will follow these steps −que1 := a queue with root0 initiallyque2 := a queue with root1 initiallywhile que1 and que2 are not empty, dotemp1 := a new list, temp2 := a new listvalues1 := a new list, values2 := a new listif que1 and que2 are containing different number of elements, thenreturn Falsefor i in range 0 ... Read More

Suppose we have a binary tree, we have to check whether for every node in the tree except leaves, its value is same as the sum of its left child's value and its right child's value or not.So, if the input is likethen the output will be TrueTo solve this, we will follow these steps −Define a function dfs() . This will take rootif root is null, thenreturn Trueif left of root is null and right of root is null, thenreturn Trueleft := 0if left of root is not null, thenleft := value of left of rootright := 0if right ... Read More

Suppose we have a list of numbers called nums and another value k, we have to find three unique entries in nums (a, b, c) such that |a + b + c − k| is minimized and return the absolute difference.So, if the input is like nums = [2, 5, 25, 6] k = 14, then the output will be 1 as if we take [2, 5, 6] will get us closest to 14 and the absolute difference is |13 − 14| = 1.To solve this, we will follow these steps −sort the list numsans := 1^9for i in range ... Read More

Suppose we have a list of numbers called nums and another value target, we have to find the number of triples (i < j < k) that exist such that nums[i] + nums[j] + nums[k] < target.So, if the input is like nums = [−2, 6, 4, 3, 8], target = 12, then the output will be 5, as the triplets are: [−2, 6, 4], [−2, 6, 3], [−2, 4, 3], [−2, 4, 8], [−2, 3, 8]To solve this, we will follow these steps −sort the list numsans := 0n := size of numsfor i in range 0 to n−1, ... Read More

Suppose we have a list of numbers called nums and another value k, we have to check whether we can find three unique elements in the list whose sum is k.So, if the input is like nums = [11, 4, 6, 10, 5, 1] k = 20, then the output will be True, as we have numbers [4, 6, 10] whose sum is 20.To solve this, we will follow these steps −sort the list numsl := 0, r := size of nums − 1while l < r − 1, dot := k − nums[l] − nums[r]if nums[r − 1] < ... Read More

Suppose we have a binary tree; we have to find the sum of values of its deepest leaves. So if the tree is like −Then the output will be 11.To solve this, we will follow these steps −Define a map m, and maxDepthDefine a recursive method solve(), this will take node and level, initially level is 0if node is not present, then returnmaxDepth := max of level and maxDepthincrease m[level] by value of nodesolve(left of node, level + 1)solve(right of node, level + 1)In the main method, setup maxDepth := 0, then solve(root, 0)return m[maxDepth]Let us see the following implementation ... Read More

Suppose we have a binary tree we have to find the sum of all right leaves in a given binary tree.So, if the input is likethen the output will be 17, as there are two right leaves in the binary tree, with values 7 and 10 respectively.To solve this, we will follow these steps −Define a function dfs(), this will take node, add, if node is null, then −returnif left of node is null and right of node is null and add is non-zero, then −ret := ret + val of nodedfs(left of node, false)dfs(right of node, true)From the main ... Read More

Suppose we have a list of numbers called nums and a value k, we have to check whether there are four unique elements in the list that add up to k.So, if the input is like nums = [11, 4, 6, 10, 5, 1] k = 25, then the output will be True, as we have [4, 6, 10, 5] whose sum is 25.To solve this, we will follow these steps −sort the list numsn := size of numsfor i in range 0 to n − 4, dofor j in range i + 1 to n − 3, dol := ... Read More