C++ Articles

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Number of Ways to Paint N × 3 Grid in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 390 Views

Suppose we have grid of size n x 3 and we want to paint each cell of the grid with exactly one of the three colors. The colors are Red, Yellow or Green. Now there is a constraint that is no two adjacent cells have the same color. We have n the number of rows of the grid. We have to find the number of ways we can paint this grid. The answer may be very large so return it modulo 10^9 + 7.So, if the input is like 1, then the output will be 12To solve this, we will ...

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Odd Even Jump in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 699 Views

Suppose we have an array A. From some starting index, we can make a series of jumps. The position (1, 3, 5, ...) jumps in the series are called odd numbered jumps, and position (2, 4, 6, ...) jumps in the series are called even numbered jumps.Now we may from index i jump forward to index j (with i < j) in this way −During odd numbered jumps, we can jump to the index j such that A[i] = A[j] and A[j] is the largest possible value. When there are multiple such indexes j, we can only jump to the ...

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Unique Paths III in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 338 Views

Suppose we have one 2-dimensional grid, there are 4 types of squares −In a square 1 is for the starting point. There will be exactly one starting square.In a square 2 is for the ending point. There will be exactly one ending square.In a square 0 is for the empty squares and we can walk over.In a square -1 if for the obstacles that we cannot walk over.We have to find the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.So, if the input is like −10000000102-1then the output ...

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Recover a Tree From Preorder Traversal in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 579 Views

Suppose there is a binary tree. We will run a preorder depth first search on the root of a binary tree.At each node in this traversal, the output will be D number of dashes (Here D is the depth of this node), after that we display the value of this node. As we know if the depth of a node is D, the depth of its immediate child is D+1 and the depth of the root node is 0.Another thing we have to keep in mind that if a node has only one child, that child is guaranteed to be ...

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Constrained Subsequence Sum in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 236 Views

Suppose we have an array called nums and an integer k, we have to find the maximum sum of a non-empty subsequence of that array such that for every two consecutive numbers in the subsequence, nums[i] and nums[j], where i < j, the condition j - i k and first element of dq is same as dp[i - k - 1], thendelete front element from dqdp[i] := maximum of dp[i] and (if dq is empty, then dp[i] + 0, otherwise first element of dp + dq[i])while (not dq is empty and last element of dq < dp[i]), do −delete ...

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Number of Valid Words for Each Puzzle in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 280 Views

Suppose there is a puzzle string, a word is valid if both the following conditions valid −word contains the first letter of puzzle.For each letter in word, that letter is in puzzle.Suppose if we consider an example that, if the puzzle is like "abcdefg", then valid words are "face", "cabbage" etc; but some invalid words are "beefed" as there is no "a" and "based" as there is "s" which is not present in the puzzle.We have to find the list of answers, where answer[i] is the number of words in the given word list words that are valid with respect ...

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Number of Ways to Stay in the Same Place After Some Steps in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 343 Views

Suppose there is an array of size arrLen, and we also have a pointer at index 0 in that array. At each step, we can move 1 position to the left, 1 position to the right in the array or stay in the same place.Now suppose we have two integers steps and arrLen, we have to find the number of ways such that the pointer still at index 0 after exactly steps. If the answer is very large then return it modulo 10^9 + 7.So, if the input is like steps = 3, arrLen = 2, then the output will ...

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Shortest Path in a Grid with Obstacles Elimination in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 697 Views

Suppose we have a m x n grid, here each cell is either 0 or 1. 0 cell is empty and 1 is blocked. In one step, we can move up, down, left or right from and to an empty cell. We have to find the minimum number of steps to walk from the upper left corner cell (0, 0) to the lower right corner cell (m-1, n-1) given that we can eliminate at most k obstacles. If there is no such way, then return -1.So, if the input is like000110000011000and k is 1, then the output will be 6, ...

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Maximum Candies You Can Get from Boxes in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 476 Views

Suppose we have n boxes, here each box is given in the format like [status, candies, keys, containedBoxes] there are some constraints −status[i]: an is 1 when box[i] is open and 0 when box[i] is closed.candies[i]: is the number of candies in box[i].keys[i]: is an array contains the indices of the boxes we can open with the key in box[i].containedBoxes[i]: an array contains the indices of the boxes found in box[i].We will start with some boxes given in initialBoxes array. We can take all the candies in any open box and we can use the keys in it to open ...

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Distinct Echo Substrings in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 231 Views

Suppose we have a string S; we have to find the number of distinct non-empty substrings of S that can be written as the concatenation of some string with itself.So, if the input is like "elloelloello", then the output will be 5, as there are some substrings like "ello", "lloe", "loel", "oell".To solve this, we will follow these steps −prime := 31m := 1^9 + 7Define a function fastPow(), this will take base, power, res := 1while power > 0, do −if power & 1 is non-zero, then −res := res * baseres := res mod mbase := base * ...

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