Count ways to split string into K Substrings starting with even digit and min length M
Last Updated :
23 Jul, 2025
Given a string s of length n consisting of digits from '0' to '9'. The task is to determine the total number of possible ways to partition the string into k substrings such that :
- Each substring has a minimum length of m (m >= 2).
- Substring must start with an even digit number and end with an odd digit number.
Examples:
Input: s = "432387429", m = 2, k = 3
Output: 3
Explanation: Following are valid partitions of the string s
43|23|87429, 4323|87|429, 43|2387|429
Input: s = "546521", m = 3, k = 2,
Output: 0
Explanation: There is no way to partition the given string.
Using Recursion - O(n^k) Time and O(k) Space
The idea is to explore all possible ways to partition the string into k substrings with minimum length m. The function starts from the beginning of the string and tries to create valid partitions by checking the digit parity conditions. At each step, it recursively explores potential partitions by moving the starting point forward and reducing the number of remaining partitions while ensuring each substring meets the specified constraints of starting with an even digit and ending with an odd digit.
Let countWays(i, k, s) represent the number of ways to partition s into k substrings starting at index i.
Base Cases:
- if length of remaining substring (s.length - i) is less than m, return 0.
- if k == 1 and the current substring is valid, return 1.
countWays(i, k, s) = countWays(j, k-1, s) where j is from i + m (to maintain minimum length of m for current substring) to last index and s[j] is an even integer and s[j-1] is an odd integer.
C++
// C++ program to count ways to
// split string into k Substrings using recursion
#include <bits/stdc++.h>
using namespace std;
int countRecur(string &s, int i, int k, int m) {
int n = s.length();
// If length of remaining substring
// is less than m, return 0.
if (n - i < m)
return 0;
// Base case for k=1.
if (k == 1)
return 1;
int ans = 0;
// Set j= i+1 to ensure length
// of current substring s[i, j]
// is atleast m.
for (int j = i + m; j < n; j++) {
int end = s[j - 1] - '0';
int start = s[j] - '0';
// If the end integer of current
// string is valid and starting
// integer of next string is valid.
if (end % 2 == 1 && start % 2 == 0) {
ans += countRecur(s, j, k - 1, m);
}
}
return ans;
}
int countWays(string &s, int m, int k) {
int start = s[0] - '0';
int n = s.length();
int end = s[n - 1] - '0';
// If the string is invalid, it cannot
// be splitted.
if (start % 2 == 1 || end % 2 == 0)
return 0;
return countRecur(s, 0, k, m);
}
int main() {
int m = 2, k = 3;
string s = "432387429";
cout << countWays(s, m, k);
return 0;
}
// Java program to count ways to
// split string into K Substrings using recursion
import java.util.*;
class GfG {
static int countRecur(String s, int i, int k, int m) {
int n = s.length();
// If length of remaining substring
// is less than m, return 0.
if (n - i < m) return 0;
// Base case for k=1.
if (k == 1) return 1;
int ans = 0;
// Set j= i+1 to ensure length
// of current substring s[i, j]
// is atleast m.
for (int j = i + m; j < n; j++) {
int end = s.charAt(j - 1) - '0';
int start = s.charAt(j) - '0';
// If the end integer of current
// string is valid and starting
// integer of next string is valid.
if (end % 2 == 1 && start % 2 == 0) {
ans += countRecur(s, j, k - 1, m);
}
}
return ans;
}
static int countWays(String s, int m, int k) {
int start = s.charAt(0) - '0';
int n = s.length();
int end = s.charAt(n - 1) - '0';
// If the string is invalid, it cannot
// be splitted.
if (start % 2 == 1 || end % 2 == 0) return 0;
return countRecur(s, 0, k, m);
}
public static void main(String[] args) {
int m = 2, k = 3;
String s = "432387429";
System.out.println(countWays(s, m, k));
}
}
# Python program to count ways to
# split string into K Substrings using recursion
def countRecur(s, i, k, m):
n = len(s)
# If length of remaining substring
# is less than m, return 0.
if n - i < m:
return 0
# Base case for k=1.
if k == 1:
return 1
ans = 0
# Set j= i+1 to ensure length
# of current substring s[i, j]
# is atleast m.
for j in range(i + m, n):
end = int(s[j - 1])
start = int(s[j])
# If the end integer of current
# string is valid and starting
# integer of next string is valid.
if end % 2 == 1 and start % 2 == 0:
ans += countRecur(s, j, k - 1, m)
return ans
def countWays(s, m, k):
start = int(s[0])
n = len(s)
end = int(s[n - 1])
# If the string is invalid, it cannot
# be splitted.
if start % 2 == 1 or end % 2 == 0:
return 0
return countRecur(s, 0, k, m)
if __name__ == "__main__":
m = 2
k = 3
s = "432387429"
print(countWays(s, m, k))
// C# program to count ways to
// split string into K Substrings using recursion
using System;
class GfG {
static int countRecur(string s, int i, int k, int m) {
int n = s.Length;
// If length of remaining substring
// is less than m, return 0.
if (n - i < m) return 0;
// Base case for k=1.
if (k == 1) return 1;
int ans = 0;
// Set j= i+1 to ensure length
// of current substring s[i, j]
// is atleast m.
for (int j = i + m; j < n; j++) {
int end = s[j - 1] - '0';
int start = s[j] - '0';
// If the end integer of current
// string is valid and starting
// integer of next string is valid.
if (end % 2 == 1 && start % 2 == 0) {
ans += countRecur(s, j, k - 1, m);
}
}
return ans;
}
static int countWays(string s, int m, int k) {
int start = s[0] - '0';
int n = s.Length;
int end = s[n - 1] - '0';
// If the string is invalid, it cannot
// be splitted.
if (start % 2 == 1 || end % 2 == 0) return 0;
return countRecur(s, 0, k, m);
}
static void Main(string[] args) {
int m = 2, k = 3;
string s = "432387429";
Console.WriteLine(countWays(s, m, k));
}
}
// JavaScript program to count ways to
// split string into K Substrings using recursion
function countRecur(s, i, k, m) {
let n = s.length;
// If length of remaining substring
// is less than m, return 0.
if (n - i < m) return 0;
// Base case for k=1.
if (k === 1) return 1;
let ans = 0;
// Set j= i+1 to ensure length
// of current substring s[i, j]
// is atleast m.
for (let j = i + m; j < n; j++) {
let end = s[j - 1] - '0';
let start = s[j] - '0';
// If the end integer of current
// string is valid and starting
// integer of next string is valid.
if (end % 2 === 1 && start % 2 === 0) {
ans += countRecur(s, j, k - 1, m);
}
}
return ans;
}
function countWays(s, m, k) {
let start = s[0] - '0';
let n = s.length;
let end = s[n - 1] - '0';
// If the string is invalid, it cannot
// be splitted.
if (start % 2 === 1 || end % 2 === 0) return 0;
return countRecur(s, 0, k, m);
}
let m = 2, k = 3;
let s = "432387429";
console.log(countWays(s, m, k));
Using Top-Down DP (Memoization) - O(n*n*k) Time and O(n*k) Space
If we notice carefully, we can observe that the above recursive solution holds the following two properties of Dynamic Programming:
1. Optimal Substructure: Number of ways to make k substrings at index i, i.e., countWays(i, k, s), depends on the optimal solutions of the subproblems countWays(j, k-1, s) where j lies between i+m and s.length. By combining these optimal substructures, we can efficiently calculate number of ways to make k substrings at index i.
2. Overlapping Subproblems: While applying a recursive approach in this problem, we notice that certain subproblems are computed multiple times.
- There are only are two parameters: i and k that changes in the recursive solution. So we create a 2D matrix of size n*(k+1) for memoization.
- We initialize this matrix as -1 to indicate nothing is computed initially.
- Now we modify our recursive solution to first check if the value is -1, then only make recursive calls. This way, we avoid re-computations of the same subproblems.
C++
// C++ program to count ways to
// split string into K Substrings using memoization
#include <bits/stdc++.h>
using namespace std;
int countRecur(string &s, int i, int k, int m,
vector<vector<int>> &memo) {
int n = s.length();
// If length of remaining substring
// is less than m, return 0.
if (n - i < m)
return 0;
// Base case for k=1.
if (k == 1)
return 1;
// If value is memoized
if (memo[i][k] != -1) {
return memo[i][k];
}
int ans = 0;
// Set j= i+1 to ensure length
// of current substring s[i, j]
// is atleast m.
for (int j = i + m; j < n; j++) {
int end = s[j - 1] - '0';
int start = s[j] - '0';
// If the end integer of current
// string is valid and starting
// integer of next string is valid.
if (end % 2 == 1 && start % 2 == 0) {
ans += countRecur(s, j, k - 1, m, memo);
}
}
return memo[i][k] = ans;
}
int countWays(string &s, int m, int k) {
int start = s[0] - '0';
int n = s.length();
int end = s[n - 1] - '0';
// If the string is invalid, it cannot
// be splitted.
if (start % 2 == 1 || end % 2 == 0)
return 0;
vector<vector<int>> memo(n, vector<int>(k + 1, -1));
return countRecur(s, 0, k, m, memo);
}
int main() {
int m = 2, k = 3;
string s = "432387429";
cout << countWays(s, m, k);
return 0;
}
// Java program to count ways to
// split string into K Substrings using memoization
import java.util.Arrays;
class GfG {
static int countRecur(String s, int i, int k,
int m, int[][] memo) {
int n = s.length();
// If length of remaining substring
// is less than m, return 0.
if (n - i < m) return 0;
// Base case for k=1.
if (k == 1) return 1;
// If value is memoized
if (memo[i][k] != -1) {
return memo[i][k];
}
int ans = 0;
// Set j= i+1 to ensure length
// of current substring s[i, j]
// is atleast m.
for (int j = i + m; j < n; j++) {
int end = s.charAt(j - 1) - '0';
int start = s.charAt(j) - '0';
// If the end integer of current
// string is valid and starting
// integer of next string is valid.
if (end % 2 == 1 && start % 2 == 0) {
ans += countRecur(s, j, k - 1, m, memo);
}
}
return memo[i][k] = ans;
}
static int countWays(String s, int m, int k) {
int start = s.charAt(0) - '0';
int n = s.length();
int end = s.charAt(n - 1) - '0';
// If the string is invalid, it cannot
// be splitted.
if (start % 2 == 1 || end % 2 == 0) return 0;
int[][] memo = new int[n][k + 1];
for (int[] row : memo) Arrays.fill(row, -1);
return countRecur(s, 0, k, m, memo);
}
public static void main(String[] args) {
int m = 2, k = 3;
String s = "432387429";
System.out.println(countWays(s, m, k));
}
}
# Python program to count ways to
# split string into K Substrings using memoization
def countRecur(s, i, k, m, memo):
n = len(s)
# If length of remaining substring
# is less than m, return 0.
if n - i < m:
return 0
# Base case for k=1.
if k == 1:
return 1
# If value is memoized
if memo[i][k] != -1:
return memo[i][k]
ans = 0
# Set j= i+1 to ensure length
# of current substring s[i, j]
# is atleast m.
for j in range(i + m, n):
end = int(s[j - 1])
start = int(s[j])
# If the end integer of current
# string is valid and starting
# integer of next string is valid.
if end % 2 == 1 and start % 2 == 0:
ans += countRecur(s, j, k - 1, m, memo)
memo[i][k] = ans
return ans
def countWays(s, m, k):
start = int(s[0])
n = len(s)
end = int(s[-1])
# If the string is invalid, it cannot
# be splitted.
if start % 2 == 1 or end % 2 == 0:
return 0
memo = [[-1] * (k + 1) for _ in range(n)]
return countRecur(s, 0, k, m, memo)
if __name__ == "__main__":
m = 2
k = 3
s = "432387429"
print(countWays(s, m, k))
// C# program to count ways to
// split string into K Substrings using memoization
using System;
class GfG {
static int countRecur(string s, int i, int k,
int m, int[,] memo) {
int n = s.Length;
// If length of remaining substring
// is less than m, return 0.
if (n - i < m) return 0;
// Base case for k=1.
if (k == 1) return 1;
// If value is memoized
if (memo[i, k] != -1) {
return memo[i, k];
}
int ans = 0;
// Set j= i+1 to ensure length
// of current substring s[i, j]
// is atleast m.
for (int j = i + m; j < n; j++) {
int end = s[j - 1] - '0';
int start = s[j] - '0';
// If the end integer of current
// string is valid and starting
// integer of next string is valid.
if (end % 2 == 1 && start % 2 == 0) {
ans += countRecur(s, j, k - 1, m, memo);
}
}
memo[i, k] = ans;
return ans;
}
static int countWays(string s, int m, int k) {
int start = s[0] - '0';
int n = s.Length;
int end = s[n - 1] - '0';
// If the string is invalid, it cannot
// be splitted.
if (start % 2 == 1 || end % 2 == 0) return 0;
int[,] memo = new int[n, k + 1];
for (int i = 0; i < n; i++)
for (int j = 0; j <= k; j++)
memo[i, j] = -1;
return countRecur(s, 0, k, m, memo);
}
static void Main(string[] args) {
int m = 2, k = 3;
string s = "432387429";
Console.WriteLine(countWays(s, m, k));
}
}
// JavaScript program to count ways to
// split string into K Substrings using memoization
function countRecur(s, i, k, m, memo) {
let n = s.length;
// If length of remaining substring
// is less than m, return 0.
if (n - i < m) return 0;
// Base case for k=1.
if (k === 1) return 1;
// If value is memoized
if (memo[i][k] !== -1) {
return memo[i][k];
}
let ans = 0;
// Set j= i+1 to ensure length
// of current substring s[i, j]
// is atleast m.
for (let j = i + m; j < n; j++) {
let end = s[j - 1] - '0';
let start = s[j] - '0';
// If the end integer of current
// string is valid and starting
// integer of next string is valid.
if (end % 2 === 1 && start % 2 === 0) {
ans += countRecur(s, j, k - 1, m, memo);
}
}
memo[i][k] = ans;
return ans;
}
function countWays(s, m, k) {
let start = s[0] - '0';
let n = s.length;
let end = s[n - 1] - '0';
// If the string is invalid, it cannot
// be splitted.
if (start % 2 === 1 || end % 2 === 0) return 0;
let memo = Array.from({ length: n }, () => Array(k + 1).fill(-1));
return countRecur(s, 0, k, m, memo);
}
let m = 2, k = 3;
let s = "432387429";
console.log(countWays(s, m, k));
Using Bottom-Up DP (Tabulation) - O(n*n*k) Time and O(n*k) Space
The idea is to fill the DP table from bottom to up. For each index i and partitions j, the dynamic programming relation is as follows:
- If string is not valid (does not start with even integer or length is less than m), set dp[i][j] = 0.
- If j == 1 and string is valid, set dp[i][j] = 1.
- Else dp[i][j] = sum(dp[x][j-1]) for all x in range [i+m, s.length].
C++
// C++ program to count ways to
// split string into K Substrings using tabulation
#include <bits/stdc++.h>
using namespace std;
int countWays(string &s, int m, int k) {
int start = s[0] - '0';
int n = s.length();
int end = s[n - 1] - '0';
// If the string is invalid, it cannot
// be splitted.
if (start % 2 == 1 || end % 2 == 0)
return 0;
vector<vector<int>> dp(n, vector<int>(k + 1, 0));
for (int i = n - 1; i >= 0; i--) {
for (int j = 1; j <= k; j++) {
int start = s[i] - '0';
// If current substring does not
// start with even digit or its
// minimum length is less than m.
if (start % 2 == 1 || n - i < m) {
dp[i][j] = 0;
continue;
}
// For k=1 and valid substring.
if (j == 1) {
dp[i][j] = 1;
continue;
}
for (int x = i + m; x < n; x++) {
int st = s[x] - '0';
int e = s[x - 1] - '0';
if (st % 2 == 0 && e % 2 == 1) {
dp[i][j] += dp[x][j - 1];
}
}
}
}
return dp[0][k];
}
int main() {
int m = 2, k = 3;
string s = "432387429";
cout << countWays(s, m, k);
return 0;
}
// Java program to count ways to
// split string into K Substrings using tabulation
class GfG {
static int countWays(String s, int m, int k) {
int start = s.charAt(0) - '0';
int n = s.length();
int end = s.charAt(n - 1) - '0';
// If the string is invalid, it cannot
// be splitted.
if (start % 2 == 1 || end % 2 == 0) return 0;
int[][] dp = new int[n][k + 1];
for (int i = n - 1; i >= 0; i--) {
for (int j = 1; j <= k; j++) {
int startDigit = s.charAt(i) - '0';
// If current substring does not
// start with even digit or its
// minimum length is less than m.
if (startDigit % 2 == 1 || n - i < m) {
dp[i][j] = 0;
continue;
}
// For k=1 and valid substring.
if (j == 1) {
dp[i][j] = 1;
continue;
}
for (int x = i + m; x < n; x++) {
int st = s.charAt(x) - '0';
int e = s.charAt(x - 1) - '0';
if (st % 2 == 0 && e % 2 == 1) {
dp[i][j] += dp[x][j - 1];
}
}
}
}
return dp[0][k];
}
public static void main(String[] args) {
int m = 2, k = 3;
String s = "432387429";
System.out.println(countWays(s, m, k));
}
}
# Python program to count ways to
# split string into K Substrings using tabulation
def countWays(s, m, k):
start = int(s[0])
n = len(s)
end = int(s[-1])
# If the string is invalid, it cannot
# be splitted.
if start % 2 == 1 or end % 2 == 0:
return 0
dp = [[0] * (k + 1) for _ in range(n)]
for i in range(n - 1, -1, -1):
for j in range(1, k + 1):
startDigit = int(s[i])
# If current substring does not
# start with even digit or its
# minimum length is less than m.
if startDigit % 2 == 1 or n - i < m:
dp[i][j] = 0
continue
# For k=1 and valid substring.
if j == 1:
dp[i][j] = 1
continue
for x in range(i + m, n):
st = int(s[x])
e = int(s[x - 1])
if st % 2 == 0 and e % 2 == 1:
dp[i][j] += dp[x][j - 1]
return dp[0][k]
if __name__ == "__main__":
m = 2
k = 3
s = "432387429"
print(countWays(s, m, k))
// C# program to count ways to
// split string into K Substrings using tabulation
using System;
class GfG {
static int countWays(string s, int m, int k) {
int start = s[0] - '0';
int n = s.Length;
int end = s[n - 1] - '0';
// If the string is invalid, it cannot
// be splitted.
if (start % 2 == 1 || end % 2 == 0) return 0;
int[,] dp = new int[n, k + 1];
for (int i = n - 1; i >= 0; i--) {
for (int j = 1; j <= k; j++) {
int startDigit = s[i] - '0';
// If current substring does not
// start with even digit or its
// minimum length is less than m.
if (startDigit % 2 == 1 || n - i < m) {
dp[i, j] = 0;
continue;
}
// For k=1 and valid substring.
if (j == 1) {
dp[i, j] = 1;
continue;
}
for (int x = i + m; x < n; x++) {
int st = s[x] - '0';
int e = s[x - 1] - '0';
if (st % 2 == 0 && e % 2 == 1) {
dp[i, j] += dp[x, j - 1];
}
}
}
}
return dp[0, k];
}
static void Main(string[] args) {
int m = 2, k = 3;
string s = "432387429";
Console.WriteLine(countWays(s, m, k));
}
}
// JavaScript program to count ways to
// split string into K Substrings using tabulation
function countWays(s, m, k) {
const start = s[0] - '0';
const n = s.length;
const end = s[n - 1] - '0';
// If the string is invalid, it cannot
// be splitted.
if (start % 2 === 1 || end % 2 === 0) return 0;
const dp = Array.from({ length: n }, () => Array(k + 1).fill(0));
for (let i = n - 1; i >= 0; i--) {
for (let j = 1; j <= k; j++) {
const startDigit = s[i] - '0';
// If current substring does not
// start with even digit or its
// minimum length is less than m.
if (startDigit % 2 === 1 || n - i < m) {
dp[i][j] = 0;
continue;
}
// For k=1 and valid substring.
if (j === 1) {
dp[i][j] = 1;
continue;
}
for (let x = i + m; x < n; x++) {
const st = s[x] - '0';
const e = s[x - 1] - '0';
if (st % 2 === 0 && e % 2 === 1) {
dp[i][j] += dp[x][j - 1];
}
}
}
}
return dp[0][k];
}
const m = 2, k = 3;
const s = "432387429";
console.log(countWays(s, m, k));
Using Space Optimized DP - O(k*n*n) Time and O(n) Space
In previous approach of dynamic programming, we observe that for calculating dp[i][j], we only need previous column dp[i][j-1]. There is no need to store all previous states.
C++
// C++ program to count ways to
// split string into K Substrings.
#include <bits/stdc++.h>
using namespace std;
int countWays(string &s, int m, int k) {
int start = s[0] - '0';
int n = s.length();
int end = s[n - 1] - '0';
// If the string is invalid, it cannot be split.
if (start % 2 == 1 || end % 2 == 0)
return 0;
vector<int> prev(n, 0), curr(n, 0);
for (int j = 1; j <= k; j++) {
for (int i = n - 1; i >= 0; i--) {
int startDigit = s[i] - '0';
// If current substring does not start with an even digit
// or its minimum length is less than m.
if (startDigit % 2 == 1 || n - i < m) {
curr[i] = 0;
continue;
}
// For k = 1 and valid substring.
if (j == 1) {
curr[i] = 1;
continue;
}
curr[i] = 0;
for (int x = i + m; x < n; x++) {
int st = s[x] - '0';
int e = s[x - 1] - '0';
if (st % 2 == 0 && e % 2 == 1) {
curr[i] += prev[x];
}
}
}
// Swap curr and prev for the next iteration.
swap(curr, prev);
}
return prev[0];
}
int main() {
int m = 2, k = 3;
string s = "432387429";
cout << countWays(s, m, k);
return 0;
}
// Java program to count ways to
// split string into K Substrings.
import java.util.*;
class GfG {
static int countWays(String s, int m, int k) {
int start = s.charAt(0) - '0';
int n = s.length();
int end = s.charAt(n - 1) - '0';
// If the string is invalid, it cannot be split.
if (start % 2 == 1 || end % 2 == 0) {
return 0;
}
int[] prev = new int[n];
int[] curr = new int[n];
for (int j = 1; j <= k; j++) {
for (int i = n - 1; i >= 0; i--) {
int startDigit = s.charAt(i) - '0';
// If current substring does not start with an even digit
// or its minimum length is less than m.
if (startDigit % 2 == 1 || n - i < m) {
curr[i] = 0;
continue;
}
// For k = 1 and valid substring.
if (j == 1) {
curr[i] = 1;
continue;
}
curr[i] = 0;
for (int x = i + m; x < n; x++) {
int st = s.charAt(x) - '0';
int e = s.charAt(x - 1) - '0';
if (st % 2 == 0 && e % 2 == 1) {
curr[i] += prev[x];
}
}
}
// Swap curr and prev for the next iteration.
int[] temp = curr;
curr = prev;
prev = temp;
}
return prev[0];
}
public static void main(String[] args) {
int m = 2, k = 3;
String s = "432387429";
System.out.println(countWays(s, m, k));
}
}
# Python program to count ways to
# split string into K Substrings.
def countWays(s, m, k):
start = int(s[0])
n = len(s)
end = int(s[n - 1])
# If the string is invalid, it cannot be split.
if start % 2 == 1 or end % 2 == 0:
return 0
prev = [0] * n
curr = [0] * n
for j in range(1, k + 1):
for i in range(n - 1, -1, -1):
startDigit = int(s[i])
# If current substring does not start with an even digit
# or its minimum length is less than m.
if startDigit % 2 == 1 or n - i < m:
curr[i] = 0
continue
# For k = 1 and valid substring.
if j == 1:
curr[i] = 1
continue
curr[i] = 0
for x in range(i + m, n):
st = int(s[x])
e = int(s[x - 1])
if st % 2 == 0 and e % 2 == 1:
curr[i] += prev[x]
# Swap curr and prev for the next iteration.
prev, curr = curr, prev
return prev[0]
if __name__ == "__main__":
m = 2
k = 3
s = "432387429"
print(countWays(s, m, k))
// C# program to count ways to
// split string into K Substrings.
using System;
class GfG {
static int countWays(string s, int m, int k) {
int start = s[0] - '0';
int n = s.Length;
int end = s[n - 1] - '0';
// If the string is invalid, it cannot be split.
if (start % 2 == 1 || end % 2 == 0) {
return 0;
}
int[] prev = new int[n];
int[] curr = new int[n];
for (int j = 1; j <= k; j++) {
for (int i = n - 1; i >= 0; i--) {
int startDigit = s[i] - '0';
// If current substring does not start with an even digit
// or its minimum length is less than m.
if (startDigit % 2 == 1 || n - i < m) {
curr[i] = 0;
continue;
}
// For k = 1 and valid substring.
if (j == 1) {
curr[i] = 1;
continue;
}
curr[i] = 0;
for (int x = i + m; x < n; x++) {
int st = s[x] - '0';
int e = s[x - 1] - '0';
if (st % 2 == 0 && e % 2 == 1) {
curr[i] += prev[x];
}
}
}
// Swap curr and prev for the next iteration.
var temp = curr;
curr = prev;
prev = temp;
}
return prev[0];
}
static void Main(string[] args) {
int m = 2, k = 3;
string s = "432387429";
Console.WriteLine(countWays(s, m, k));
}
}
// JavaScript program to count ways to
// split string into K Substrings.
function countWays(s, m, k) {
const start = parseInt(s[0]);
const n = s.length;
const end = parseInt(s[n - 1]);
// If the string is invalid, it cannot be split.
if (start % 2 === 1 || end % 2 === 0) {
return 0;
}
let prev = new Array(n).fill(0);
let curr = new Array(n).fill(0);
for (let j = 1; j <= k; j++) {
for (let i = n - 1; i >= 0; i--) {
const startDigit = parseInt(s[i]);
// If current substring does not start with an even digit
// or its minimum length is less than m.
if (startDigit % 2 === 1 || n - i < m) {
curr[i] = 0;
continue;
}
// For k = 1 and valid substring.
if (j === 1) {
curr[i] = 1;
continue;
}
curr[i] = 0;
for (let x = i + m; x < n; x++) {
const st = parseInt(s[x]);
const e = parseInt(s[x - 1]);
if (st % 2 === 0 && e % 2 === 1) {
curr[i] += prev[x];
}
}
}
// Swap curr and prev for the next iteration.
[curr, prev] = [prev, curr];
}
return prev[0];
}
const m = 2, k = 3;
const s = "432387429";
console.log(countWays(s, m, k));
Time Complexity: O(k*n*n), due to the 3 nested for loops: outer loop (from 1 to k), inner loop ( n-1 to 0 ) , innermost loop ( In the worst case, the innermost loop executes about n - i iterations, which is roughly n for each iteration of i)
Space Complexity: O(n), both vectors prev and curr takes O(n) space as they store values for each index of the string.
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