Maximum string length after choosing strings from given Array with given conditions
Last Updated :
05 Apr, 2023
Given an array of string S[] of size N, the task is to find the maximum size of the resultant string formed by adding some strings and following the given condition that If a string of size K is chosen to add in the resultant string then the next K/2 strings cannot be selected to be a part of the resultant array.
Examples:
Input: S[] = {"well", "do", "hi", "by"}
Output: 6
Explanation: Choose "well" and skip "do" and "hi"(sizeof("well")/2) and then choose "by". So, size will be 6.
Input: s[] = {"geeks", "for", "geeks", "is", "best"}
Output: 9
Approach: This problem can be solved using memoization. Follow the steps below:
- For each string S[i], there are two options i.e. to choose the current string or not.
- So if the string is chosen, its length, say K will contribute to the length of the resultant array and now, only the strings after K/2 can be chosen.
- Now if the string is excluded, just move further.
- Print the answer according to the above observation
Below is the implementation of the above approach
C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
// Recursive function to find the
// maximum length of the resultant string
int maxsum(string S[], int N, int i,
vector<int>& dp)
{
// If index gets out of bound return 0
if (i >= N)
return 0;
// If value not already computed
// then compute it
if (dp[i] == -1) {
// To include the current string
int op1
= S[i].size()
+ maxsum(S, N,
(i + S[i].size() / 2)
+ 1,
dp);
// To exclude the current string
int op2 = maxsum(S, N, i + 1, dp);
// Maximum of both the options
dp[i] = max(op1, op2);
}
return dp[i];
}
// Driver Code
int main()
{
string S[] = { "geeks", "for", "geeks",
"is", "best" };
int N = sizeof(S) / sizeof(S[0]);
vector<int> dp(N, -1);
cout << maxsum(S, N, 0, dp);
return 0;
}
// Java implementation of the above approach
import java.util.Arrays;
class GFG {
// Recursive function to find the
// maximum length of the resultant string
static int maxsum(String S[], int N, int i, int[] dp)
{
// If index gets out of bound return 0
if (i >= N)
return 0;
// If value not already computed
// then compute it
if (dp[i] == -1) {
// To include the current string
int op1 = S[i].length()
+ maxsum(S, N,
(i + S[i].length() / 2)
+ 1,
dp);
// To exclude the current string
int op2 = maxsum(S, N, i + 1, dp);
// Maximum of both the options
dp[i] = Math.max(op1, op2);
}
return dp[i];
}
// Driver Code
public static void main(String args[]) {
String S[] = { "geeks", "for", "geeks", "is", "best" };
int N = S.length;
int[] dp = new int[N];
Arrays.fill(dp, -1);
System.out.println(maxsum(S, N, 0, dp));
}
}
// This code is contributed by saurabh_jaiswal.
# Python implementation of the above approach
# Recursive function to find the
# maximum length of the resultant string
def maxsum(S, N, i, dp):
# If index gets out of bound return 0
if (i >= N):
return 0
# If value not already computed
# then compute it
if (dp[i] == -1):
# To include the current string
op1 = int(len(S[i]) + maxsum(S, N, (i + len(S[i]) // 2)+1, dp))
# To exclude the current string
op2 = int(maxsum(S, N, i + 1, dp))
# Maximum of both the options
dp[i] = max(op1, op2)
return dp[i]
# Driver Code
S = ["geeks", "for", "geeks", "is", "best"]
N = len(S)
dp = []
for i in range(0, N):
dp.append(-1)
print(maxsum(S, N, 0, dp))
# This code is contributed by Taranpreet
// C# implementation of the above approach
using System;
public class GFG
{
// Recursive function to find the
// maximum length of the resultant string
static int maxsum(String []S, int N, int i, int[] dp)
{
// If index gets out of bound return 0
if (i >= N)
return 0;
// If value not already computed
// then compute it
if (dp[i] == -1) {
// To include the current string
int op1 = S[i].Length + maxsum(S, N, (i + S[i].Length / 2) + 1, dp);
// To exclude the current string
int op2 = maxsum(S, N, i + 1, dp);
// Maximum of both the options
dp[i] = Math.Max(op1, op2);
}
return dp[i];
}
// Driver Code
public static void Main(String []args)
{
String []S = { "geeks", "for", "geeks", "is", "best" };
int N = S.Length;
int[] dp = new int[N];
for(int i = 0;i<N;i++)
dp[i] = -1;
Console.WriteLine(maxsum(S, N, 0, dp));
}
}
// This code is contributed by umadevi9616
<script>
// JavaScript code for the above approach
// Recursive function to find the
// maximum length of the resultant string
function maxsum(S, N, i,
dp) {
// If index gets out of bound return 0
if (i >= N)
return 0;
// If value not already computed
// then compute it
if (dp[i] == -1) {
// To include the current string
let op1
= S[i].length
+ maxsum(S, N,
(i + Math.floor(S[i].length / 2))
+ 1,
dp);
// To exclude the current string
let op2 = maxsum(S, N, i + 1, dp);
// Maximum of both the options
dp[i] = Math.max(op1, op2);
}
return dp[i];
}
// Driver Code
let S = ["geeks", "for", "geeks",
"is", "best"];
let N = S.length;
let dp = new Array(N).fill(-1)
document.write(maxsum(S, N, 0, dp));
// This code is contributed by Potta Lokesh
</script>
Time Complexity: O(N)
Auxiliary Space: O(N)
Efficient approach : Using DP Tabulation method ( Iterative approach )
The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memorization(top-down) because memorization method needs extra stack space of recursion calls.
Steps to solve this problem :
- Create a vector to store the solution of the subproblems.
- Initialize the table with base cases
- Fill up the table iteratively
- Return the final solution
Implementation :
C++
// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
// Iterative function to find the
// maximum length of the resultant string
int maxsum(string S[], int N)
{
// create vector to store the computations of
// subproblems
vector<int> dp(N + 1);
dp[N] = 0; // base case
// loop to compute the bigger values
for (int i = N - 1; i >= 0; i--) {
int op1 = S[i].size() + dp[i + S[i].size() / 2 + 1];
int op2 = dp[i + 1];
// store answer of subproblem
dp[i] = max(op1, op2);
}
// return answer
return dp[0];
}
// Driver code
int main()
{
string S[] = { "geeks", "for", "geeks", "is", "best" };
int N = sizeof(S) / sizeof(S[0]);
cout << maxsum(S, N) << endl;
return 0;
}
// this code is contributed by bhardwajji
import java.util.*;
class Main {
// Iterative function to find the maximum length of the
// resultant string
static int maxsum(String S[], int N)
{
// create array to store the computations of
// subproblems
int[] dp = new int[N + 1];
dp[N] = 0; // base case
// loop to compute the bigger values
for (int i = N - 1; i >= 0; i--) {
int j = i + S[i].length() / 2 + 1;
int op1 = j < N ? S[i].length() + dp[j]
: S[i].length();
int op2 = dp[i + 1];
// store answer of subproblem
dp[i] = Math.max(op1, op2);
}
// return answer
return dp[0];
}
// Driver code
public static void main(String[] args)
{
String S[]
= { "geeks", "for", "geeks", "is", "best" };
int N = S.length;
System.out.println(maxsum(S, N));
}
}
def maxsum(S, N):
# create list to store the computations of subproblems
dp = [0] * (N + 1)
dp[N] = 0 # base case
# loop to compute the bigger values
for i in range(N - 1, -1, -1):
op1 = len(S[i]) + dp[min(i + len(S[i]) // 2 + 1, N)]
op2 = dp[i + 1]
# store answer of subproblem
dp[i] = max(op1, op2)
# return answer
return dp[0]
# Driver code
S = ["geeks", "for", "geeks", "is", "best"]
N = len(S)
print(maxsum(S, N))
// C# program for above approach
using System;
using System.Linq;
using System.Collections.Generic;
public class Program {
// Iterative function to find the
// maximum length of the resultant string
public static int MaxSum(string[] S, int N)
{
// create list to store the computations of
// subproblems
List<int> dp = new List<int>(new int[N + 1]);
dp[N] = 0; // base case
// loop to compute the bigger values
for (int i = N - 1; i >= 0; i--) {
int op1 = S[i].Length
+ (i + S[i].Length / 2 + 1 <= N
? dp[i + S[i].Length / 2 + 1]
: 0);
int op2 = dp[i + 1];
// store answer of subproblem
dp[i] = Math.Max(op1, op2);
}
// return answer
return dp[0];
}
// Driver code
public static void Main()
{
string[] S
= { "geeks", "for", "geeks", "is", "best" };
int N = S.Length;
Console.WriteLine(MaxSum(S, N));
}
}
// this code is contributed by chetanbargal
function maxsum(S, N) {
// create array to store the computations of subproblems
let dp = new Array(N + 1).fill(0);
dp[N] = 0; // base case
// loop to compute the bigger values
for (let i = N - 1; i >= 0; i--) {
let op1 = S[i].length + dp[Math.min(i + Math.floor(S[i].length / 2) + 1, N)];
let op2 = dp[i + 1];
// store answer of subproblem
dp[i] = Math.max(op1, op2);
}
// return answer
return dp[0];
}
// Driver code
let S = ["geeks", "for", "geeks", "is", "best"];
let N = S.length;
console.log(maxsum(S, N));
// This code is contributed by user_dtewbxkn77n
Output :
9
Time Complexity: O(N)
Auxiliary Space: O(N)
Similar Reads
DSA Tutorial - Learn Data Structures and Algorithms DSA (Data Structures and Algorithms) is the study of organizing data efficiently using data structures like arrays, stacks, and trees, paired with step-by-step procedures (or algorithms) to solve problems effectively. Data structures manage how data is stored and accessed, while algorithms focus on
7 min read
Quick Sort QuickSort is a sorting algorithm based on the Divide and Conquer that picks an element as a pivot and partitions the given array around the picked pivot by placing the pivot in its correct position in the sorted array. It works on the principle of divide and conquer, breaking down the problem into s
12 min read
Merge Sort - Data Structure and Algorithms Tutorials Merge sort is a popular sorting algorithm known for its efficiency and stability. It follows the divide-and-conquer approach. It works by recursively dividing the input array into two halves, recursively sorting the two halves and finally merging them back together to obtain the sorted array. Merge
14 min read
Bubble Sort Algorithm Bubble Sort is the simplest sorting algorithm that works by repeatedly swapping the adjacent elements if they are in the wrong order. This algorithm is not suitable for large data sets as its average and worst-case time complexity are quite high.We sort the array using multiple passes. After the fir
8 min read
Data Structures Tutorial Data structures are the fundamental building blocks of computer programming. They define how data is organized, stored, and manipulated within a program. Understanding data structures is very important for developing efficient and effective algorithms. What is Data Structure?A data structure is a st
2 min read
Breadth First Search or BFS for a Graph Given a undirected graph represented by an adjacency list adj, where each adj[i] represents the list of vertices connected to vertex i. Perform a Breadth First Search (BFS) traversal starting from vertex 0, visiting vertices from left to right according to the adjacency list, and return a list conta
15+ min read
Binary Search Algorithm - Iterative and Recursive Implementation Binary Search Algorithm is a searching algorithm used in a sorted array by repeatedly dividing the search interval in half. The idea of binary search is to use the information that the array is sorted and reduce the time complexity to O(log N). Binary Search AlgorithmConditions to apply Binary Searc
15 min read
Insertion Sort Algorithm Insertion sort is a simple sorting algorithm that works by iteratively inserting each element of an unsorted list into its correct position in a sorted portion of the list. It is like sorting playing cards in your hands. You split the cards into two groups: the sorted cards and the unsorted cards. T
9 min read
Dijkstra's Algorithm to find Shortest Paths from a Source to all Given a weighted undirected graph represented as an edge list and a source vertex src, find the shortest path distances from the source vertex to all other vertices in the graph. The graph contains V vertices, numbered from 0 to V - 1.Note: The given graph does not contain any negative edge. Example
12 min read
Selection Sort Selection Sort is a comparison-based sorting algorithm. It sorts an array by repeatedly selecting the smallest (or largest) element from the unsorted portion and swapping it with the first unsorted element. This process continues until the entire array is sorted.First we find the smallest element an
8 min read