Minimum cost to sort strings using reversals of different costs
Last Updated :
19 Apr, 2025
Given an array of strings arr[] and an array cost[] representing the cost to reverse each corresponding string. We are not allowed to move or reorder the strings in the array, we can only reverse individual strings if needed. The task is to choose which strings to reverse such that the final array is in non-decreasing lexicographic order, and the total cost of reversals is minimized.
Note: If it's not possible to achieve a sorted array using any combination of reversals, return -1.
Examples:
Input: arr[] = ["aa", "ba", "ac"], cost[] = [1, 3, 1]
Output: 1
Explanation: The minimum cost to make the array sorted lexicographically is 1.
-> "aa" at index 0 is fine, whether reversed or not, as "aa" remains the same.
-> "ba" at index 1 is greater than "aa", so it's already in correct order.
-> "ac" at index 2 is less than "ba", so the array is not sorted. Reversing "ac" gives "ca", which is greater than "ba", making the array ["aa", "ba", "ca"] sorted. This single reversal costs 1, and no other reversal option gives a better result.
Input: arr[] = ["cba", "bca", "aaa"], cost[] = [1, 2, 3]
Output: -1
Explanation: No combination of reversals can make the array sorted in lexicographic order. Even reversing all strings results in ["abc", "acb", "aaa"], which is also not sorted. Hence, the output is -1.
[Approach 1] Using Dynamic Programming - O(n) Time and O(n) Space
At each index i, we have two choices:
- Keep
arr[i] as-is - Reverse
arr[i] (with cost cost[i])
We recursively decide which choice to make, such that:
- The chosen version of
arr[i] is lexicographically ≥ the previous chosen string. - We minimize the total cost.
The idea is to use dynamic programming. We maintain a 2D dp table where dp[i][0] means the minimum cost to keep i-th string as it is and dp[i][1] means the cost if it's reversed. The observation is that a string can only follow the previous one in the list if it's lexicographically greater than or equal to it (either reversed or not). We try both cases at every index and update the cost only if the order is maintained, and finally return the minimum of both states at the last index.
C++
// C++ program to get minimum cost to sort
// strings by reversal operation using DP
#include <bits/stdc++.h>
using namespace std;
// Function to return minimum cost to make array sorted
// using only string reversals. Returns -1 if not possible.
int minCost(vector<string> arr, vector<int> cost) {
int n = arr.size();
// dp[i][0] = min cost if i-th string is not reversed
// dp[i][1] = min cost if i-th string is reversed
vector<vector<int>> dp(n, vector<int>(2, INT_MAX));
// First string: cost 0 if kept, cost[0] if reversed
dp[0][0] = 0;
dp[0][1] = cost[0];
// Store reversed versions of strings
vector<string> reversed(n);
for (int i = 0; i < n; i++) {
reversed[i] = arr[i];
reverse(reversed[i].begin(), reversed[i].end());
}
// Fill dp for all strings from second to last
for (int i = 1; i < n; i++) {
// j = 0 means the string stri[i] is reversed
// j = 1 means not reversed
for (int j = 0; j <= 1; j++) {
// Current string and cost if reversed or not
string currStr = j ? reversed[i] : arr[i];
int currCost = j ? cost[i] : 0;
// If currStr >= previous without reverse
if (currStr >= arr[i - 1] && dp[i - 1][0] != INT_MAX) {
dp[i][j] = min(dp[i][j],
dp[i - 1][0] + currCost);
}
// If currStr >= previous with reverse
if (currStr >= reversed[i - 1] && dp[i - 1][1] != INT_MAX) {
dp[i][j] = min(dp[i][j],
dp[i - 1][1] + currCost);
}
}
}
// Get min of last string reversed or not
int res = min(dp[n - 1][0], dp[n - 1][1]);
return (res == INT_MAX) ? -1 : res;
}
// Driver code
int main() {
vector<string> arr = {"aa", "ba", "ac"};
vector<int> cost = {1, 3, 1};
cout << minCost(arr, cost);
return 0;
}
// Java program to get minimum cost to sort
// strings by reversal operation using DP
import java.util.*;
class GfG {
// Function to return minimum cost to make array sorted
// using only string reversals. Returns -1 if not possible.
static int minCost(String[] arr, int[] cost) {
int n = arr.length;
// dp[i][0] = min cost if i-th string is not reversed
// dp[i][1] = min cost if i-th string is reversed
int[][] dp = new int[n][2];
for (int[] row : dp)
Arrays.fill(row, Integer.MAX_VALUE);
// First string: cost 0 if kept, cost[0] if reversed
dp[0][0] = 0;
dp[0][1] = cost[0];
// Store reversed versions of strings
String[] reversed = new String[n];
for (int i = 0; i < n; i++) {
reversed[i] = new StringBuilder(arr[i]).reverse().toString();
}
// Fill dp for all strings from second to last
for (int i = 1; i < n; i++) {
// j = 0 means the string stri[i] is reversed
// j = 1 means not reversed
for (int j = 0; j <= 1; j++) {
// Current string and cost if reversed or not
String currStr = (j == 1) ? reversed[i] : arr[i];
int currCost = (j == 1) ? cost[i] : 0;
// If currStr >= previous without reverse
if (currStr.compareTo(arr[i - 1]) >= 0 && dp[i - 1][0] != Integer.MAX_VALUE) {
dp[i][j] = Math.min(dp[i][j],
dp[i - 1][0] + currCost);
}
// If currStr >= previous with reverse
if (currStr.compareTo(reversed[i - 1]) >= 0 && dp[i - 1][1] != Integer.MAX_VALUE) {
dp[i][j] = Math.min(dp[i][j],
dp[i - 1][1] + currCost);
}
}
}
// Get min of last string reversed or not
int res = Math.min(dp[n - 1][0], dp[n - 1][1]);
return (res == Integer.MAX_VALUE) ? -1 : res;
}
public static void main(String[] args) {
String[] arr = {"aa", "ba", "ac"};
int[] cost = {1, 3, 1};
System.out.println(minCost(arr, cost));
}
}
import sys
def minCost(arr, cost):
n = len(arr)
# Initialize dp table with max values
dp = [[sys.maxsize] * 2 for _ in range(n)]
# First string: cost 0 if not reversed, cost[0] if reversed
dp[0][0] = 0
dp[0][1] = cost[0]
# Precompute reversed strings
reversed_arr = [s[::-1] for s in arr]
for i in range(1, n):
# j = 0 means not reversed and 1 means reversed
for j in range(2):
curr_str = reversed_arr[i] if j == 1 else arr[i]
curr_cost = cost[i] if j == 1 else 0
# Compare with previous not reversed
if curr_str >= arr[i - 1] and dp[i - 1][0] != sys.maxsize:
dp[i][j] = min(dp[i][j], dp[i - 1][0] + curr_cost)
# Compare with previous reversed
if curr_str >= reversed_arr[i - 1] and dp[i - 1][1] != sys.maxsize:
dp[i][j] = min(dp[i][j], dp[i - 1][1] + curr_cost)
result = min(dp[n - 1][0], dp[n - 1][1])
return -1 if result == sys.maxsize else result
# Driver code
if __name__ == "__main__":
arr = ["aa", "ba", "ac"]
cost = [1, 3, 1]
print(minCost(arr, cost))
// C# program to get minimum cost to sort
// strings by reversal operation using DP
using System;
using System.Linq;
class GfG {
// Function to return minimum cost to make array sorted
// using only string reversals. Returns -1 if not possible.
static int MinCost(string[] arr, int[] cost) {
int n = arr.Length;
// dp[i][0] = min cost if i-th string is not reversed
// dp[i][1] = min cost if i-th string is reversed
int[,] dp = new int[n, 2];
for (int i = 0; i < n; i++)
dp[i, 0] = dp[i, 1] = int.MaxValue;
// First string: cost 0 if kept, cost[0] if reversed
dp[0, 0] = 0;
dp[0, 1] = cost[0];
// Store reversed versions of strings
string[] reversed = new string[n];
for (int i = 0; i < n; i++) {
char[] charArray = arr[i].ToCharArray();
Array.Reverse(charArray);
reversed[i] = new string(charArray);
}
// Fill dp for all strings from second to last
for (int i = 1; i < n; i++) {
// j = 0 means the string stri[i] is reversed
// j = 1 means not reversed
for (int j = 0; j <= 1; j++) {
// Current string and cost if reversed or not
string currStr = (j == 1) ? reversed[i] : arr[i];
int currCost = (j == 1) ? cost[i] : 0;
// If currStr >= previous without reverse
if (string.Compare(currStr, arr[i - 1]) >= 0 && dp[i - 1, 0] != int.MaxValue) {
dp[i, j] = Math.Min(dp[i, j], dp[i - 1, 0] + currCost);
}
// If currStr >= previous with reverse
if (string.Compare(currStr, reversed[i - 1]) >= 0 && dp[i - 1, 1] != int.MaxValue) {
dp[i, j] = Math.Min(dp[i, j], dp[i - 1, 1] + currCost);
}
}
}
// Get min of last string reversed or not
int res = Math.Min(dp[n - 1, 0], dp[n - 1, 1]);
return (res == int.MaxValue) ? -1 : res;
}
public static void Main(string[] args) {
string[] arr = {"aa", "ba", "ac"};
int[] cost = {1, 3, 1};
Console.WriteLine(MinCost(arr, cost));
}
}
// Importing required modules
function minCost(arr, cost) {
const n = arr.length;
// Initialize dp table with max values
const dp = Array.from({ length: n }, () => [Infinity, Infinity]);
// First string: cost 0 if not reversed, cost[0] if reversed
dp[0][0] = 0;
dp[0][1] = cost[0];
// Precompute reversed strings
const reversedArr = arr.map(s => s.split('').reverse().join(''));
for (let i = 1; i < n; i++) {
// j = 0 means not reversed and 1 means reversed
for (let j = 0; j < 2; j++) {
const currStr = j === 1 ? reversedArr[i] : arr[i];
const currCost = j === 1 ? cost[i] : 0;
// Compare with previous not reversed
if (currStr >= arr[i - 1] && dp[i - 1][0] !== Infinity) {
dp[i][j] = Math.min(dp[i][j], dp[i - 1][0] + currCost);
}
// Compare with previous reversed
if (currStr >= reversedArr[i - 1] && dp[i - 1][1] !== Infinity) {
dp[i][j] = Math.min(dp[i][j], dp[i - 1][1] + currCost);
}
}
}
const result = Math.min(dp[n - 1][0], dp[n - 1][1]);
return result === Infinity ? -1 : result;
}
// Driver code
const arr = ["aa", "ba", "ac"];
const cost = [1, 3, 1];
console.log(minCost(arr, cost));
[Approach 2] Using Space Optimized DP - O(n) Time and O(1) Space
The idea is to reduce space complexity by avoiding the use of a full 2D dp array. Instead of storing the minimum costs for all indices, we only track the previous state using two variables, which is sufficient for the current computation. This optimization brings down the space complexity from O(n) to O(1).
Steps to implement the above idea:
- Initialize dp0 and dp1 to track minimum cost for original and reversed first string respectively.
- Loop through the array from index 1 to n-1 and compute cur0 and cur1 as INT_MAX initially.
- For each string, update cur0 if it's lexicographically greater than previous original or reversed string.
- Reverse current and previous strings to compute cur1 based on comparisons with original and reversed forms.
- Add cost[i] while updating cur1 since reversal incurs a cost in those valid transitions.
- Update dp0 and dp1 to current cur0 and cur1 respectively for the next iteration.
- Return the minimum of dp0 and dp1, or -1 if both are still INT_MAX, indicating it's unsortable.
C++
// C++ program to get minimum cost to sort
// strings by reversal operation using
// Space Optimized DP
#include <bits/stdc++.h>
using namespace std;
// Returns minimum cost for sorting arr
// using reverse operation. This function
// returns -1 if it is not possible to sort.
int minCost(vector<string> arr, vector<int> cost) {
int n = arr.size();
// Minimum cost for the previous string
// in original order
int dp0 = 0;
// Minimum cost for the previous
// string in reversed order
int dp1 = cost[0];
for (int i = 1; i < n; i++) {
// Minimum cost for the current
// string in original order
int cur0 = INT_MAX;
// Minimum cost for the current
// string in reversed order
int cur1 = INT_MAX;
// Update dp values only if the current string is
// lexicographically larger
if (arr[i] >= arr[i - 1]) {
cur0 = min(cur0, dp0);
}
if (arr[i] >= string(arr[i - 1].rbegin(), arr[i - 1].rend())) {
cur0 = min(cur0, dp1);
}
// Update dp values for reversed strings
string revCurr = string(arr[i].rbegin(), arr[i].rend());
string revPrev = string(arr[i - 1].rbegin(), arr[i - 1].rend());
if (revCurr >= arr[i - 1]) {
cur1 = min(cur1, dp0 + cost[i]);
}
if (revCurr >= revPrev) {
cur1 = min(cur1, dp1 + cost[i]);
}
// Update the minimum cost for
// the previous string
dp0 = cur0;
dp1 = cur1;
}
// Get the minimum from both entries
// of the last index
int res = min(dp0, dp1);
return (res == INT_MAX) ? -1 : res;
}
// Driver code
int main() {
vector<string> arr = {"aa", "ba", "ac"};
vector<int> cost = {1, 3, 1};
cout << minCost(arr, cost);;
return 0;
}
// Java program to get minimum cost to sort
// strings by reversal operation using
// Space Optimized DP
class GfG {
// Returns minimum cost for sorting arr
// using reverse operation. This function
// returns -1 if it is not possible to sort.
public static int minCost(String[] arr, int[] cost) {
int n = arr.length;
// Minimum cost for the previous string
// in original order
int dp0 = 0;
// Minimum cost for the previous
// string in reversed order
int dp1 = cost[0];
for (int i = 1; i < n; i++) {
// Minimum cost for the current
// string in original order
int cur0 = Integer.MAX_VALUE;
// Minimum cost for the current
// string in reversed order
int cur1 = Integer.MAX_VALUE;
// Update dp values only if the current string is
// lexicographically larger
if (arr[i].compareTo(arr[i - 1]) >= 0) {
cur0 = Math.min(cur0, dp0);
}
String revPrev = new StringBuilder(arr[i - 1]).reverse().toString();
if (arr[i].compareTo(revPrev) >= 0) {
cur0 = Math.min(cur0, dp1);
}
// Update dp values for reversed strings
String revCurr = new StringBuilder(arr[i]).reverse().toString();
if (revCurr.compareTo(arr[i - 1]) >= 0) {
cur1 = Math.min(cur1, dp0 + cost[i]);
}
if (revCurr.compareTo(revPrev) >= 0) {
cur1 = Math.min(cur1, dp1 + cost[i]);
}
// Update the minimum cost for
// the previous string
dp0 = cur0;
dp1 = cur1;
}
// Get the minimum from both entries
// of the last index
int res = Math.min(dp0, dp1);
return (res == Integer.MAX_VALUE) ? -1 : res;
}
public static void main(String[] args) {
String[] arr = {"aa", "ba", "ac"};
int[] cost = {1, 3, 1};
System.out.println(minCost(arr, cost));
}
}
# Python program to get minimum cost to sort
# strings by reversal operation using
# Space Optimized DP
# Returns minimum cost for sorting arr
# using reverse operation. This function
# returns -1 if it is not possible to sort.
def minCost(arr, cost):
n = len(arr)
# Minimum cost for the previous string
# in original order
dp0 = 0
# Minimum cost for the previous
# string in reversed order
dp1 = cost[0]
for i in range(1, n):
# Minimum cost for the current
# string in original order
cur0 = float('inf')
# Minimum cost for the current
# string in reversed order
cur1 = float('inf')
# Update dp values only if the current string is
# lexicographically larger
if arr[i] >= arr[i - 1]:
cur0 = min(cur0, dp0)
if arr[i] >= arr[i - 1][::-1]:
cur0 = min(cur0, dp1)
# Update dp values for reversed strings
revCurr = arr[i][::-1]
revPrev = arr[i - 1][::-1]
if revCurr >= arr[i - 1]:
cur1 = min(cur1, dp0 + cost[i])
if revCurr >= revPrev:
cur1 = min(cur1, dp1 + cost[i])
# Update the minimum cost for
# the previous string
dp0 = cur0
dp1 = cur1
# Get the minimum from both entries
# of the last index
res = min(dp0, dp1)
return -1 if res == float('inf') else res
if __name__ == "__main__":
arr = ["aa", "ba", "ac"]
cost = [1, 3, 1]
print(minCost(arr, cost))
// C# program to get minimum cost to sort
// strings by reversal operation using
// Space Optimized DP
using System;
class GfG {
// Returns minimum cost for sorting arr
// using reverse operation. This function
// returns -1 if it is not possible to sort.
public static int minCost(string[] arr, int[] cost) {
int n = arr.Length;
// Minimum cost for the previous string
// in original order
int dp0 = 0;
// Minimum cost for the previous
// string in reversed order
int dp1 = cost[0];
for (int i = 1; i < n; i++) {
// Minimum cost for the current
// string in original order
int cur0 = int.MaxValue;
// Minimum cost for the current
// string in reversed order
int cur1 = int.MaxValue;
// Update dp values only if the current string is
// lexicographically larger
if (String.Compare(arr[i], arr[i - 1]) >= 0) {
cur0 = Math.Min(cur0, dp0);
}
string revPrev = new string(arr[i - 1].ToCharArray().Reverse().ToArray());
if (String.Compare(arr[i], revPrev) >= 0) {
cur0 = Math.Min(cur0, dp1);
}
// Update dp values for reversed strings
string revCurr = new string(arr[i].ToCharArray().Reverse().ToArray());
if (String.Compare(revCurr, arr[i - 1]) >= 0) {
cur1 = Math.Min(cur1, dp0 + cost[i]);
}
if (String.Compare(revCurr, revPrev) >= 0) {
cur1 = Math.Min(cur1, dp1 + cost[i]);
}
// Update the minimum cost for
// the previous string
dp0 = cur0;
dp1 = cur1;
}
// Get the minimum from both entries
// of the last index
int res = Math.Min(dp0, dp1);
return (res == int.MaxValue) ? -1 : res;
}
public static void Main() {
string[] arr = {"aa", "ba", "ac"};
int[] cost = {1, 3, 1};
Console.WriteLine(minCost(arr, cost));
}
}
// JavaScript program to get minimum cost to sort
// strings by reversal operation using
// Space Optimized DP
// Returns minimum cost for sorting arr
// using reverse operation. This function
// returns -1 if it is not possible to sort.
function minCost(arr, cost) {
let n = arr.length;
// Minimum cost for the previous string
// in original order
let dp0 = 0;
// Minimum cost for the previous
// string in reversed order
let dp1 = cost[0];
for (let i = 1; i < n; i++) {
// Minimum cost for the current
// string in original order
let cur0 = Infinity;
// Minimum cost for the current
// string in reversed order
let cur1 = Infinity;
// Update dp values only if the current string is
// lexicographically larger
if (arr[i] >= arr[i - 1]) {
cur0 = Math.min(cur0, dp0);
}
if (arr[i] >= arr[i - 1].split('').reverse().join('')) {
cur0 = Math.min(cur0, dp1);
}
// Update dp values for reversed strings
let revCurr = arr[i].split('').reverse().join('');
let revPrev = arr[i - 1].split('').reverse().join('');
if (revCurr >= arr[i - 1]) {
cur1 = Math.min(cur1, dp0 + cost[i]);
}
if (revCurr >= revPrev) {
cur1 = Math.min(cur1, dp1 + cost[i]);
}
// Update the minimum cost for
// the previous string
dp0 = cur0;
dp1 = cur1;
}
// Get the minimum from both entries
// of the last index
let res = Math.min(dp0, dp1);
return (res === Infinity) ? -1 : res;
}
// Driver code
let arr = ["aa", "ba", "ac"];
let cost = [1, 3, 1];
console.log(minCost(arr, cost));
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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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