Given integers n (the height of the tower), m (the maximum size of tiles available), and k (the maximum number of times each tile size can be used), the task is to calculate the number of distinct stable towers of height n that can be built.
Note:
- A stable tower consists of exactly n tiles, each stacked such that no larger tile is placed on a smaller one.
- Two towers are considered different if, at any height h (where 1<=h<=n), they have tiles of different sizes.
Examples:
Input: n = 3, m = 3, k = 1.
Output: 1
Explanation: Possible sequence: [1, 2, 3].
Input: n = 3, m = 3, k = 2.
Output: 7
Explanation: Possible sequences: [1, 1, 2], [1, 1, 3], [1, 2, 2], [1, 2, 3], [1, 3, 3], [2, 2, 3], [2, 3, 3].
Using Recursion - O(k^m) Time and O(m) Space
The main idea for the recursive approach is to build the tower by selecting different numbers of tiles from the largest available tile size, moving to smaller sizes as we progress. For each tile size, we try adding up to k tiles (or fewer, if fewer are needed to reach height n). By reducing both the remaining height n and the largest available tile size m with each recursive call, we explore all valid combinations that maintain the stability condition (no larger tile placed on a smaller one).
The recurrence relation for this solution can be expressed as:
possibleWays(n, m, k) = sum of possibleWays(n - cnt, m - 1, k) for each cnt from 0 to min(k, n), where possibleWays(n, m, k) gives the number of ways to build a tower of height n using up to size m tiles, with a maximum of k tiles of any size allowed.
Base cases:
- ways(0, m, k) = 1, meaning a tower of height 0 has only one possible configuration (using no tiles).
- ways(n, 0, k) = 0, meaning no tower can be built if m is zero and n is greater than zero.
C++
// C++ program to find number of ways to
// make stable towers of given height.
#include <bits/stdc++.h>
using namespace std;
// Recursive function to calculate possible ways to
// form towers of height 'n' using tiles of sizes
// up to 'm' with at most 'k' tiles of each size
int possibleWaysRecur(int n, int m, int k) {
// Base cases:
// If height is 0, there is exactly 1 way (empty tower)
if (n == 0) return 1;
// If no more tile sizes are available,
// no ways can be formed
if (m == 0) return 0;
int ans = 0;
// Try using 'cnt' tiles of the current size
// (from 0 up to 'k' tiles) and make recursive
// calls for remaining height and tiles
for (int cnt = 0; cnt <= k && cnt <= n; cnt++) {
ans += possibleWaysRecur(n - cnt, m - 1, k);
}
return ans;
}
int possibleWays(int n, int m, int k) {
return possibleWaysRecur(n, m, k);
}
int main() {
int n = 3, m = 3, k = 2;
cout << possibleWays(n, m, k) << endl;
return 0;
}
// Java program to find number of ways to
// make stable towers of given height.
class GfG {
static int possibleWaysRecur(int n, int m, int k) {
// Base cases:
// If height is 0, there is exactly 1 way (empty tower)
if (n == 0) return 1;
// If no more tile sizes are available,
// no ways can be formed
if (m == 0) return 0;
int ans = 0;
// Try using 'cnt' tiles of the current size
// (from 0 up to 'k' tiles) and make recursive
// calls for remaining height and tiles
for (int cnt = 0; cnt <= k && cnt <= n; cnt++) {
ans += possibleWaysRecur(n - cnt, m - 1, k);
}
return ans;
}
static int possibleWays(int n, int m, int k) {
return possibleWaysRecur(n, m, k);
}
public static void main(String[] args) {
int n = 3, m = 3, k = 2;
System.out.println(possibleWays(n, m, k));
}
}
# Python program to find number of ways to
# make stable towers of given height.
def possibleWaysRecur(n, m, k):
# Base cases:
# If height is 0, there is exactly 1 way (empty tower)
if n == 0:
return 1
# If no more tile sizes are available,
# no ways can be formed
if m == 0:
return 0
ans = 0
# Try using 'cnt' tiles of the current size
# (from 0 up to 'k' tiles) and make recursive
# calls for remaining height and tiles
for cnt in range(min(k, n) + 1):
ans += possibleWaysRecur(n - cnt, m - 1, k)
return ans
def possibleWays(n, m, k):
return possibleWaysRecur(n, m, k)
if __name__ == "__main__":
n, m, k = 3, 3, 2
print(possibleWays(n, m, k))
// C# program to find number of ways to
// make stable towers of given height.
using System;
class GfG {
static int possibleWaysRecur(int n, int m, int k) {
// Base cases:
// If height is 0, there is exactly 1 way (empty tower)
if (n == 0) return 1;
// If no more tile sizes are available,
// no ways can be formed
if (m == 0) return 0;
int ans = 0;
// Try using 'cnt' tiles of the current size
// (from 0 up to 'k' tiles) and make recursive
// calls for remaining height and tiles
for (int cnt = 0; cnt <= k && cnt <= n; cnt++) {
ans += possibleWaysRecur(n - cnt, m - 1, k);
}
return ans;
}
static int possibleWays(int n, int m, int k) {
return possibleWaysRecur(n, m, k);
}
static void Main() {
int n = 3, m = 3, k = 2;
Console.WriteLine(possibleWays(n, m, k));
}
}
// JavaScript program to find number of ways to
// make stable towers of given height.
function possibleWaysRecur(n, m, k) {
// Base cases:
// If height is 0, there is exactly 1 way (empty tower)
if (n === 0) return 1;
// If no more tile sizes are available,
// no ways can be formed
if (m === 0) return 0;
let ans = 0;
// Try using 'cnt' tiles of the current size
// (from 0 up to 'k' tiles) and make recursive
// calls for remaining height and tiles
for (let cnt = 0; cnt <= k && cnt <= n; cnt++) {
ans += possibleWaysRecur(n - cnt, m - 1, k);
}
return ans;
}
function possibleWays(n, m, k) {
return possibleWaysRecur(n, m, k);
}
const n = 3, m = 3, k = 2;
console.log(possibleWays(n, m, k));
Using Top-Down DP (Memoization) - O(n*m*k) Time and O(n*m) Space
If we notice carefully, we can observe that the above recursive solution holds the following two properties of Dynamic Programming.
1. Optimal Substructure: The solution to the tile stacking problem can be derived from the optimal solutions of smaller subproblems. Specifically, for any given n (height of tower) and m (sizes of tiles), we can express the recursive relation as follows:
- possibleWays(n, m, k) = sum of possibleWays(n - cnt, m - 1, k) for each cnt from 0 to min(k, n).
2. Overlapping Subproblems: When implementing a recursive approach to solve the tile stacking problem, we observe that many subproblems are computed multiple times.
- The recursive solution involves changing two parameters: n (height of tower) and the current tile size (m). We need to track both parameters, so we create a 2D array of size (n+1) x (m+1) because the value of n will be in the range [0, n] and m will be in the range [0, m+1].
- We initialize the 2D array with -1 to indicate that no subproblems have been computed yet.
- We check if the value at memo[n][m] is -1. If it is, we proceed to compute the result. otherwise, we return the stored result.
C++
// C++ program to find number of ways to
// make stable towers of given height.
#include <bits/stdc++.h>
using namespace std;
// Recursive function to calculate possible ways to
// form towers of height 'n' using tiles of sizes
// up to 'm' with at most 'k' tiles of each size
int possibleWaysRecur(int n, int m, int k,
vector<vector<int>> &memo) {
// Base cases:
// If height is 0, there is exactly 1 way (empty tower)
if (n == 0) return 1;
// If no more tile sizes are available,
// no ways can be formed
if (m == 0) return 0;
if (memo[n][m] != -1) return memo[n][m];
int ans = 0;
// Try using 'cnt' tiles of the current size
// (from 0 up to 'k' tiles) and make recursive
// calls for remaining height and tiles
for (int cnt = 0; cnt <= k && cnt <= n; cnt++) {
ans += possibleWaysRecur(n - cnt, m - 1, k, memo);
}
return memo[n][m] = ans;
}
int possibleWays(int n, int m, int k) {
vector<vector<int>> memo(n + 1, vector<int>(m + 1, -1));
return possibleWaysRecur(n, m, k, memo);
}
int main() {
int n = 3, m = 3, k = 2;
cout << possibleWays(n, m, k) << endl;
return 0;
}
// Java program to find number of ways to
// make stable towers of given height.
class GfG {
static int possibleWaysRecur(int n, int m, int k, int[][] memo) {
// Base cases:
// If height is 0, there is exactly 1 way (empty tower)
if (n == 0) return 1;
// If no more tile sizes are available,
// no ways can be formed
if (m == 0) return 0;
if (memo[n][m] != -1) return memo[n][m];
int ans = 0;
// Try using 'cnt' tiles of the current size
// (from 0 up to 'k' tiles) and make recursive
// calls for remaining height and tiles
for (int cnt = 0; cnt <= k && cnt <= n; cnt++) {
ans += possibleWaysRecur(n - cnt, m - 1, k, memo);
}
return memo[n][m] = ans;
}
static int possibleWays(int n, int m, int k) {
int[][] memo = new int[n + 1][m + 1];
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
memo[i][j] = -1;
}
}
return possibleWaysRecur(n, m, k, memo);
}
public static void main(String[] args) {
int n = 3, m = 3, k = 2;
System.out.println(possibleWays(n, m, k));
}
}
# Python program to find number of ways to
# make stable towers of given height.
def possibleWaysRecur(n, m, k, memo):
# Base cases:
# If height is 0, there is exactly 1 way (empty tower)
if n == 0:
return 1
# If no more tile sizes are available,
# no ways can be formed
if m == 0:
return 0
if memo[n][m] != -1:
return memo[n][m]
ans = 0
# Try using 'cnt' tiles of the current size
# (from 0 up to 'k' tiles) and make recursive
# calls for remaining height and tiles
for cnt in range(min(k, n) + 1):
ans += possibleWaysRecur(n - cnt, m - 1, k, memo)
memo[n][m] = ans
return ans
def possibleWays(n, m, k):
memo = [[-1 for _ in range(m + 1)] for _ in range(n + 1)]
return possibleWaysRecur(n, m, k, memo)
if __name__ == "__main__":
n, m, k = 3, 3, 2
print(possibleWays(n, m, k))
// C# program to find number of ways to
// make stable towers of given height.
using System;
class GfG {
static int possibleWaysRecur(int n, int m, int k, int[,] memo) {
// Base cases:
// If height is 0, there is exactly 1 way (empty tower)
if (n == 0) return 1;
// If no more tile sizes are available,
// no ways can be formed
if (m == 0) return 0;
if (memo[n, m] != -1) return memo[n, m];
int ans = 0;
// Try using 'cnt' tiles of the current size
// (from 0 up to 'k' tiles) and make recursive
// calls for remaining height and tiles
for (int cnt = 0; cnt <= k && cnt <= n; cnt++) {
ans += possibleWaysRecur(n - cnt, m - 1, k, memo);
}
memo[n, m] = ans;
return ans;
}
static int possibleWays(int n, int m, int k) {
int[,] memo = new int[n + 1, m + 1];
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
memo[i, j] = -1;
}
}
return possibleWaysRecur(n, m, k, memo);
}
static void Main() {
int n = 3, m = 3, k = 2;
Console.WriteLine(possibleWays(n, m, k));
}
}
// JavaScript program to find number of ways to
// make stable towers of given height.
function possibleWaysRecur(n, m, k, memo) {
// Base cases:
// If height is 0, there is exactly 1 way (empty tower)
if (n === 0) return 1;
// If no more tile sizes are available,
// no ways can be formed
if (m === 0) return 0;
if (memo[n][m] !== -1) return memo[n][m];
let ans = 0;
// Try using 'cnt' tiles of the current size
// (from 0 up to 'k' tiles) and make recursive
// calls for remaining height and tiles
for (let cnt = 0; cnt <= k && cnt <= n; cnt++) {
ans += possibleWaysRecur(n - cnt, m - 1, k, memo);
}
memo[n][m] = ans;
return ans;
}
function possibleWays(n, m, k) {
let memo = Array.from({ length: n + 1 }, () => Array(m + 1).fill(-1));
return possibleWaysRecur(n, m, k, memo);
}
const n = 3, m = 3, k = 2;
console.log(possibleWays(n, m, k));
Using Bottom-Up DP (Tabulation) - O(n*m*k) Time and O(m*n) Space
The approach is similar to the previous one. Just instead of breaking down the problem recursively, we iteratively build up the solution by calculating in bottom-up manner.
The idea is to build the dp table where dp[i][j] represents the number of ways to form a tower of height i using tiles of sizes from 1 to j, with each tile being used at most k times.
The table is initialized with dp[0][j] = 1 for all j, as there is exactly one way to form a tower of height 0 (using no tiles). For each i (height) and j (tile size), the number of ways to form the tower is computed by first considering the ways without using the jth tile (dp[i][j] = dp[i][j-1]), and then adding the possibilities of using the jth tile from 1 to k times, provided the height i allows it (dp[i][j] += dp[i - cnt][j - 1] for each valid cnt).
C++
// C++ program to find number of ways to
// make stable towers of given height.
#include <bits/stdc++.h>
using namespace std;
// function to calculate possible ways to form
// towers of height 'n' using tiles of sizes up
// to 'm' with at most 'k' tiles of each size
int possibleWays(int n, int m, int k) {
vector<vector<int>> dp(n + 1, vector<int>(m + 1, 0));
// Base case for height 0
for (int j = 0; j <= m; j++) dp[0][j] = 1;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
// Not using the jth tile at all
dp[i][j] = dp[i][j - 1];
for (int cnt = 1; cnt <= k && cnt <= i; cnt++) {
dp[i][j] += dp[i - cnt][j - 1];
}
}
}
return dp[n][m];
}
int main() {
int n = 3, m = 3, k = 2;
cout << possibleWays(n, m, k) << endl;
return 0;
}
// Java program to find number of ways to
// make stable towers of given height.
class GfG {
static int possibleWays(int n, int m, int k) {
int[][] dp = new int[n + 1][m + 1];
// Base case for height 0
for (int j = 0; j <= m; j++) dp[0][j] = 1;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
// Not using the jth tile at all
dp[i][j] = dp[i][j - 1];
for (int cnt = 1; cnt <= k && cnt <= i; cnt++) {
dp[i][j] += dp[i - cnt][j - 1];
}
}
}
return dp[n][m];
}
public static void main(String[] args) {
int n = 3, m = 3, k = 2;
System.out.println(possibleWays(n, m, k));
}
}
# Python program to find number of ways to
# make stable towers of given height.
def possibleWays(n, m, k):
dp = [[0 for _ in range(m + 1)] for _ in range(n + 1)]
# Base case for height 0
for j in range(m + 1):
dp[0][j] = 1
for i in range(1, n + 1):
for j in range(1, m + 1):
# Not using the jth tile at all
dp[i][j] = dp[i][j - 1]
for cnt in range(1, min(k, i) + 1):
dp[i][j] += dp[i - cnt][j - 1]
return dp[n][m]
if __name__ == "__main__":
n, m, k = 3, 3, 2
print(possibleWays(n, m, k))
// C# program to find number of ways to
// make stable towers of given height.
using System;
class GfG {
static int possibleWays(int n, int m, int k) {
int[,] dp = new int[n + 1, m + 1];
// Base case for height 0
for (int j = 0; j <= m; j++) dp[0, j] = 1;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
// Not using the jth tile at all
dp[i, j] = dp[i, j - 1];
for (int cnt = 1; cnt <= k && cnt <= i; cnt++) {
dp[i, j] += dp[i - cnt, j - 1];
}
}
}
return dp[n, m];
}
static void Main() {
int n = 3, m = 3, k = 2;
Console.WriteLine(possibleWays(n, m, k));
}
}
// JavaScript program to find number of ways to
// make stable towers of given height.
function possibleWays(n, m, k) {
let dp = Array.from({ length: n + 1 }, () => Array(m + 1).fill(0));
// Base case for height 0
for (let j = 0; j <= m; j++) dp[0][j] = 1;
for (let i = 1; i <= n; i++) {
for (let j = 1; j <= m; j++) {
// Not using the jth tile at all
dp[i][j] = dp[i][j - 1];
for (let cnt = 1; cnt <= k && cnt <= i; cnt++) {
dp[i][j] += dp[i - cnt][j - 1];
}
}
}
return dp[n][m];
}
const n = 3, m = 3, k = 2;
console.log(possibleWays(n, m, k));
Using Space Optimized DP - O(m*n*k) Time and O(n) Space
From the previous approach, we can observe that that value of dp[i][j] depends only on the previous row, ie, i-1. There is no need to store all the previous states just one previous state is used to compute result.
C++
// C++ program to find number of ways to
// make stable towers of given height.
#include <bits/stdc++.h>
using namespace std;
int possibleWays(int n, int m, int k) {
vector<int> curr(n + 1, 0), prev(n + 1, 0);
// Base case for height 0
prev[0] = 1;
for (int j = 1; j <= m; j++) {
// Base case for height 0 in the current row
curr[0] = 1;
for (int i = 1; i <= n; i++) {
curr[i] = prev[i];
for (int cnt = 1; cnt <= k && cnt <= i; cnt++) {
curr[i] += prev[i - cnt];
}
}
swap(curr, prev);
}
return prev[n];
}
int main() {
int n = 3, m = 3, k = 2;
cout << possibleWays(n, m, k) << endl;
return 0;
}
// Java program to find number of ways to
// make stable towers of given height.
class GfG {
static int possibleWays(int n, int m, int k) {
int[] curr = new int[n + 1];
int[] prev = new int[n + 1];
// Base case for height 0
prev[0] = 1;
for (int j = 1; j <= m; j++) {
// Base case for height 0 in the current row
curr[0] = 1;
for (int i = 1; i <= n; i++) {
curr[i] = prev[i];
for (int cnt = 1; cnt <= k && cnt <= i; cnt++) {
curr[i] += prev[i - cnt];
}
}
// Swap curr and prev arrays
int[] temp = curr;
curr = prev;
prev = temp;
}
return prev[n];
}
public static void main(String[] args) {
int n = 3, m = 3, k = 2;
System.out.println(possibleWays(n, m, k));
}
}
# Python program to find number of ways to
# make stable towers of given height.
def possibleWays(n, m, k):
curr = [0] * (n + 1)
prev = [0] * (n + 1)
# Base case for height 0
prev[0] = 1
for j in range(1, m + 1):
# Base case for height 0 in the current row
curr[0] = 1
for i in range(1, n + 1):
curr[i] = prev[i]
for cnt in range(1, min(k, i) + 1):
curr[i] += prev[i - cnt]
# Swap curr and prev arrays
prev, curr = curr, prev
return prev[n]
if __name__ == "__main__":
n, m, k = 3, 3, 2
print(possibleWays(n, m, k))
// C# program to find number of ways to
// make stable towers of given height.
using System;
class GfG {
static int possibleWays(int n, int m, int k) {
int[] curr = new int[n + 1];
int[] prev = new int[n + 1];
// Base case for height 0
prev[0] = 1;
for (int j = 1; j <= m; j++) {
// Base case for height 0 in the current row
curr[0] = 1;
for (int i = 1; i <= n; i++) {
curr[i] = prev[i];
for (int cnt = 1; cnt <= k && cnt <= i; cnt++) {
curr[i] += prev[i - cnt];
}
}
// Swap curr and prev arrays
int[] temp = curr;
curr = prev;
prev = temp;
}
return prev[n];
}
static void Main() {
int n = 3, m = 3, k = 2;
Console.WriteLine(possibleWays(n, m, k));
}
}
// JavaScript program to find number of ways to
// make stable towers of given height.
function possibleWays(n, m, k) {
let curr = new Array(n + 1).fill(0);
let prev = new Array(n + 1).fill(0);
// Base case for height 0
prev[0] = 1;
for (let j = 1; j <= m; j++) {
// Base case for height 0 in the current row
curr[0] = 1;
for (let i = 1; i <= n; i++) {
curr[i] = prev[i];
for (let cnt = 1; cnt <= k && cnt <= i; cnt++) {
curr[i] += prev[i - cnt];
}
}
// Swap curr and prev arrays
[curr, prev] = [prev, curr];
}
return prev[n];
}
const n = 3, m = 3, k = 2;
console.log(possibleWays(n, m, k));
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Coin Change - Count Ways to Make SumGiven an integer array of coins[] of size n representing different types of denominations and an integer sum, the task is to count all combinations of coins to make a given value sum. Note: Assume that you have an infinite supply of each type of coin. Examples: Input: sum = 4, coins[] = [1, 2, 3]Out
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Subset Sum ProblemGiven an array arr[] of non-negative integers and a value sum, the task is to check if there is a subset of the given array whose sum is equal to the given sum. Examples: Input: arr[] = [3, 34, 4, 12, 5, 2], sum = 9Output: TrueExplanation: There is a subset (4, 5) with sum 9.Input: arr[] = [3, 34, 4
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Introduction and Dynamic Programming solution to compute nCr%pGiven three numbers n, r and p, compute value of nCr mod p. Example: Input: n = 10, r = 2, p = 13 Output: 6 Explanation: 10C2 is 45 and 45 % 13 is 6.We strongly recommend that you click here and practice it, before moving on to the solution.METHOD 1: (Using Dynamic Programming) A Simple Solution is
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Rod CuttingGiven a rod of length n inches and an array price[]. price[i] denotes the value of a piece of length i. The task is to determine the maximum value obtainable by cutting up the rod and selling the pieces.Note: price[] is 1-indexed array.Input: price[] = [1, 5, 8, 9, 10, 17, 17, 20]Output: 22Explanati
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Painting Fence AlgorithmGiven a fence with n posts and k colors, the task is to find out the number of ways of painting the fence so that not more than two consecutive posts have the same color.Examples:Input: n = 2, k = 4Output: 16Explanation: We have 4 colors and 2 posts.Ways when both posts have same color: 4 Ways when
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Longest Common Subsequence (LCS)Given two strings, s1 and s2, the task is to find the length of the Longest Common Subsequence. If there is no common subsequence, return 0. A subsequence is a string generated from the original string by deleting 0 or more characters, without changing the relative order of the remaining characters.
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Longest Increasing Subsequence (LIS)Given an array arr[] of size n, the task is to find the length of the Longest Increasing Subsequence (LIS) i.e., the longest possible subsequence in which the elements of the subsequence are sorted in increasing order.Examples: Input: arr[] = [3, 10, 2, 1, 20]Output: 3Explanation: The longest increa
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Longest subsequence such that difference between adjacents is oneGiven an array arr[] of size n, the task is to find the longest subsequence such that the absolute difference between adjacent elements is 1.Examples: Input: arr[] = [10, 9, 4, 5, 4, 8, 6]Output: 3Explanation: The three possible subsequences of length 3 are [10, 9, 8], [4, 5, 4], and [4, 5, 6], wher
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Maximum size square sub-matrix with all 1sGiven a binary matrix mat of size n * m, the task is to find out the maximum length of a side of a square sub-matrix with all 1s.Example:Input: mat = [ [0, 1, 1, 0, 1], [1, 1, 0, 1, 0], [0, 1, 1, 1, 0], [1, 1, 1, 1, 0], [1, 1, 1, 1, 1], [0, 0, 0, 0, 0] ]Output: 3Explanation: The maximum length of a
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Min Cost PathYou are given a 2D matrix cost[][] of dimensions m × n, where each cell represents the cost of traversing through that position. Your goal is to determine the minimum cost required to reach the bottom-right cell (m-1, n-1) starting from the top-left cell (0,0).The total cost of a path is the sum of
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Longest Common Substring (Space optimized DP solution)Given two strings ‘s1‘ and ‘s2‘, find the length of the longest common substring. Example: Input: s1 = “GeeksforGeeksâ€, s2 = “GeeksQuiz†Output : 5 Explanation:The longest common substring is “Geeks†and is of length 5.Input: s1 = “abcdxyzâ€, s2 = “xyzabcd†Output : 4Explanation:The longest common su
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Count ways to reach the nth stair using step 1, 2 or 3A child is running up a staircase with n steps and can hop either 1 step, 2 steps, or 3 steps at a time. The task is to implement a method to count how many possible ways the child can run up the stairs.Examples: Input: 4Output: 7Explanation: There are seven ways: {1, 1, 1, 1}, {1, 2, 1}, {2, 1, 1},
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Grid Unique Paths - Count Paths in matrixGiven an matrix of size m x n, the task is to find the count of all unique possible paths from top left to the bottom right with the constraints that from each cell we can either move only to the right or down.Examples: Input: m = 2, n = 2Output: 2Explanation: There are two paths(0, 0) -> (0, 1)
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Unique paths in a Grid with ObstaclesGiven a matrix mat[][] of size n * m, where mat[i][j] = 1 indicates an obstacle and mat[i][j] = 0 indicates an empty space. The task is to find the number of unique paths to reach (n-1, m-1) starting from (0, 0). You are allowed to move in the right or downward direction. Note: In the grid, cells ma
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Medium problems on Dynamic programming
0/1 Knapsack ProblemGiven n items where each item has some weight and profit associated with it and also given a bag with capacity W, [i.e., the bag can hold at most W weight in it]. The task is to put the items into the bag such that the sum of profits associated with them is the maximum possible. Note: The constraint
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Printing Items in 0/1 KnapsackGiven weights and values of n items, put these items in a knapsack of capacity W to get the maximum total value in the knapsack. In other words, given two integer arrays, val[0..n-1] and wt[0..n-1] represent values and weights associated with n items respectively. Also given an integer W which repre
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Unbounded Knapsack (Repetition of items allowed)Given a knapsack weight, say capacity and a set of n items with certain value vali and weight wti, The task is to fill the knapsack in such a way that we can get the maximum profit. This is different from the classical Knapsack problem, here we are allowed to use an unlimited number of instances of
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Egg Dropping Puzzle | DP-11You are given n identical eggs and you have access to a k-floored building from 1 to k.There exists a floor f where 0 <= f <= k such that any egg dropped from a floor higher than f will break, and any egg dropped from or below floor f will not break. There are a few rules given below:An egg th
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Word BreakGiven a string s and y a dictionary of n words dictionary, check if s can be segmented into a sequence of valid words from the dictionary, separated by spaces.Examples:Input: s = "ilike", dictionary[] = ["i", "like", "gfg"]Output: trueExplanation: The string can be segmented as "i like".Input: s = "
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Vertex Cover Problem (Dynamic Programming Solution for Tree)A vertex cover of an undirected graph is a subset of its vertices such that for every edge (u, v) of the graph, either ‘u’ or ‘v’ is in vertex cover. Although the name is Vertex Cover, the set covers all edges of the given graph. The problem to find minimum size vertex cover of a graph is NP complet
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Tile Stacking ProblemGiven integers n (the height of the tower), m (the maximum size of tiles available), and k (the maximum number of times each tile size can be used), the task is to calculate the number of distinct stable towers of height n that can be built. Note:A stable tower consists of exactly n tiles, each stac
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Box Stacking ProblemGiven three arrays height[], width[], and length[] of size n, where height[i], width[i], and length[i] represent the dimensions of a box. The task is to create a stack of boxes that is as tall as possible, but we can only stack a box on top of another box if the dimensions of the 2-D base of the low
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Partition a Set into Two Subsets of Equal SumGiven an array arr[], the task is to check if it can be partitioned into two parts such that the sum of elements in both parts is the same.Note: Each element is present in either the first subset or the second subset, but not in both.Examples: Input: arr[] = [1, 5, 11, 5]Output: true Explanation: Th
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Travelling Salesman Problem using Dynamic ProgrammingGiven a 2d matrix cost[][] of size n where cost[i][j] denotes the cost of moving from city i to city j. The task is to complete a tour from city 0 (0-based index) to all other cities such that we visit each city exactly once and then at the end come back to city 0 at minimum cost.Note the difference
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Longest Palindromic Subsequence (LPS)Given a string s, find the length of the Longest Palindromic Subsequence in it. Note: The Longest Palindromic Subsequence (LPS) is the maximum-length subsequence of a given string that is also a Palindrome. Longest Palindromic SubsequenceExamples:Input: s = "bbabcbcab"Output: 7Explanation: Subsequen
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Longest Common Increasing Subsequence (LCS + LIS)Given two arrays, a[] and b[], find the length of the longest common increasing subsequence(LCIS). LCIS refers to a subsequence that is present in both arrays and strictly increases.Prerequisites: LCS, LIS.Examples:Input: a[] = [3, 4, 9, 1], b[] = [5, 3, 8, 9, 10, 2, 1]Output: 2Explanation: The long
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Find all distinct subset (or subsequence) sums of an arrayGiven an array arr[] of size n, the task is to find a distinct sum that can be generated from the subsets of the given sets and return them in increasing order. It is given that the sum of array elements is small.Examples: Input: arr[] = [1, 2]Output: [0, 1, 2, 3]Explanation: Four distinct sums can
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Weighted Job SchedulingGiven a 2D array jobs[][] of order n*3, where each element jobs[i] defines start time, end time, and the profit associated with the job. The task is to find the maximum profit you can take such that there are no two jobs with overlapping time ranges.Note: If the job ends at time X, it is allowed to
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Count Derangements (Permutation such that no element appears in its original position)A Derangement is a permutation of n elements, such that no element appears in its original position. For example, a derangement of [0, 1, 2, 3] is [2, 3, 1, 0].Given a number n, find the total number of Derangements of a set of n elements.Examples : Input: n = 2Output: 1Explanation: For two balls [1
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Minimum insertions to form a palindromeGiven a string s, the task is to find the minimum number of characters to be inserted to convert it to a palindrome.Examples:Input: s = "geeks"Output: 3Explanation: "skgeegks" is a palindromic string, which requires 3 insertions.Input: s= "abcd"Output: 3Explanation: "abcdcba" is a palindromic string
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Ways to arrange Balls such that adjacent balls are of different typesThere are 'p' balls of type P, 'q' balls of type Q and 'r' balls of type R. Using the balls we want to create a straight line such that no two balls of the same type are adjacent.Examples : Input: p = 1, q = 1, r = 0Output: 2Explanation: There are only two arrangements PQ and QPInput: p = 1, q = 1,
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