Given a large number in form of string str of size N, the task is to count the subsequence of length 4 whose digit are in the form of (x, x, x+1, x+1).
Example:
Input: str = “1515732322”
Output: 3
Explanation:
For the given input string str = “1515732322”, there are 3 subsequence {1122}, {1122}, and {1122} which are in the given form of (x, x, x+1, x+1).Input: str = “224353”
Output: 1
Explanation:
For the given input string str = “224353”, there is only 1 subsequence possible {2233} in the given form of (x, x, x+1, x+1).
Prefix Sum Approach: Please refer to the Set 1, for the Prefix sum approach.
Dynamic Programming Approach: This problem can be solved using Dynamic Programming.
We will be using 2 arrays as count1[][j] and count2[][10] such that count1[i][10] will store the count of consecutive equal element of digit j at current index i traversing string from left and count2[i][j] will store the count of consecutive equal element of digit j at current index i traversing string from right. Below are the steps:
- Initialize two count array count1[][10] for filling table from left to right and count2[][10] for filling table from right to left of input string.
- Traverse input string and fill the both count1 and count2 array.
- Recurrence Relation for count1[][] is given by:
count1[i][j] += count1[i – 1][j]
where count1[i][j] is the count of two adjacent at index i for digit j
- Recurrence Relation for count2[][] is given by:
count2[i][j] += count1[i + 1][j]
where count2[i][j] is the count of two adjacent at index i for digit j
- Initialize a variable ans to 0 that stores the resultant count of stable numbers.
- Traverse the input string and get the count of numbers from count1[][] and count2[][] array such that the difference between number from count1[][] and count2[][] array is 1 and store it in variable c1 and c2.
- Finally update result(say ans) with (c1 * ((c2 * (c2 – 1) / 2))).
- Print the answer ans calculated above.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;
// Function to count the numbersint countStableNum(string str, int N)
{ // Array that stores the
// digits from left to right
int count1[N][10];
// Array that stores the
// digits from right to left
int count2[N][10];
// Initially both array store zero
for (int i = 0; i < N; i++)
for (int j = 0; j < 10; j++)
count1[i][j] = count2[i][j] = 0;
// Fill the table for count1 array
for (int i = 0; i < N; i++) {
if (i != 0) {
for (int j = 0; j < 10; j++) {
count1[i][j] += count1[i - 1][j];
}
}
// Update the count of current character
count1[i][str[i] - '0']++;
}
// Fill the table for count2 array
for (int i = N - 1; i >= 0; i--) {
if (i != N - 1) {
for (int j = 0; j < 10; j++) {
count2[i][j] += count2[i + 1][j];
}
}
// Update the count of current character
count2[i][str[i] - '0']++;
}
// Variable that stores the
// count of the numbers
int ans = 0;
// Traverse Input string and get the
// count of digits from count1 and
// count2 array such that difference
// b/w digit is 1 & store it int c1 &c2.
// And store it in variable c1 and c2
for (int i = 1; i < N - 1; i++) {
if (str[i] == '9')
continue;
int c1 = count1[i - 1][str[i] - '0'];
int c2 = count2[i + 1][str[i] - '0' + 1];
if (c2 == 0)
continue;
// Update the ans
ans = (ans
+ (c1 * ((c2 * (c2 - 1) / 2))));
}
// Return the final count
return ans;
}// Driver Codeint main()
{ // Given String
string str = "224353";
int N = str.length();
// Function Call
cout << countStableNum(str, N);
return 0;
} |
Java
// Java program for the above approachimport java.io.*;
class GFG{
// Function to count the numbersstatic int countStableNum(String str, int N)
{ // Array that stores the
// digits from left to right
int count1[][] = new int[N][10];
// Array that stores the
// digits from right to left
int count2[][] = new int[N][10];
// Initially both array store zero
for(int i = 0; i < N; i++)
for(int j = 0; j < 10; j++)
count1[i][j] = count2[i][j] = 0;
// Fill the table for count1 array
for(int i = 0; i < N; i++)
{
if (i != 0)
{
for(int j = 0; j < 10; j++)
{
count1[i][j] += count1[i - 1][j];
}
}
// Update the count of current character
count1[i][str.charAt(i) - '0']++;
}
// Fill the table for count2 array
for(int i = N - 1; i >= 0; i--)
{
if (i != N - 1)
{
for(int j = 0; j < 10; j++)
{
count2[i][j] += count2[i + 1][j];
}
}
// Update the count of current character
count2[i][str.charAt(i) - '0']++;
}
// Variable that stores the
// count of the numbers
int ans = 0;
// Traverse Input string and get the
// count of digits from count1 and
// count2 array such that difference
// b/w digit is 1 & store it int c1 &c2.
// And store it in variable c1 and c2
for(int i = 1; i < N - 1; i++)
{
if (str.charAt(i) == '9')
continue;
int c1 = count1[i - 1][str.charAt(i) - '0'];
int c2 = count2[i + 1][str.charAt(i) - '0' + 1];
if (c2 == 0)
continue;
// Update the ans
ans = (ans + (c1 * ((c2 * (c2 - 1) / 2))));
}
// Return the final count
return ans;
}// Driver codepublic static void main(String[] args)
{ // Given String
String str = "224353";
int N = str.length();
// Function call
System.out.println(countStableNum(str, N));
}}// This code is contributed by Pratima Pandey |
Python3
# Python3 program for the above approach # Function to count the numbers def countStableNum(Str, N):
# Array that stores the
# digits from left to right
count1 = [[0 for j in range(10)]
for i in range(N)]
# Array that stores the
# digits from right to left
count2 = [[0 for j in range(10)]
for i in range(N)]
# Initially both array store zero
for i in range(N):
for j in range(10):
count1[i][j], count2[i][j] = 0, 0
# Fill the table for count1 array
for i in range(N):
if (i != 0):
for j in range(10):
count1[i][j] = (count1[i][j] +
count1[i - 1][j])
# Update the count of current character
count1[i][ord(Str[i]) - ord('0')] += 1
# Fill the table for count2 array
for i in range(N - 1, -1, -1):
if (i != N - 1):
for j in range(10):
count2[i][j] += count2[i + 1][j]
# Update the count of current character
count2[i][ord(Str[i]) -
ord('0')] = count2[i][ord(Str[i]) -
ord('0')] + 1
# Variable that stores the
# count of the numbers
ans = 0
# Traverse Input string and get the
# count of digits from count1 and
# count2 array such that difference
# b/w digit is 1 & store it int c1 &c2.
# And store it in variable c1 and c2
for i in range(1, N - 1):
if (Str[i] == '9'):
continue
c1 = count1[i - 1][ord(Str[i]) - ord('0')]
c2 = count2[i + 1][ord(Str[i]) - ord('0') + 1]
if (c2 == 0):
continue
# Update the ans
ans = (ans + (c1 * ((c2 * (c2 - 1) // 2))))
# Return the final count
return ans
# Driver code# Given String Str = "224353"
N = len(Str)
# Function call print(countStableNum(Str, N))
# This code is contributed by divyeshrabadiya07 |
C#
// C# program for the above approachusing System;
class GFG{
// Function to count the numbersstatic int countStableNum(String str, int N)
{ // Array that stores the
// digits from left to right
int [,]count1 = new int[N, 10];
// Array that stores the
// digits from right to left
int [,]count2 = new int[N, 10];
// Initially both array store zero
for(int i = 0; i < N; i++)
for(int j = 0; j < 10; j++)
count1[i, j] = count2[i, j] = 0;
// Fill the table for count1 array
for(int i = 0; i < N; i++)
{
if (i != 0)
{
for(int j = 0; j < 10; j++)
{
count1[i, j] += count1[i - 1, j];
}
}
// Update the count of current character
count1[i, str[i] - '0']++;
}
// Fill the table for count2 array
for(int i = N - 1; i >= 0; i--)
{
if (i != N - 1)
{
for(int j = 0; j < 10; j++)
{
count2[i, j] += count2[i + 1, j];
}
}
// Update the count of current character
count2[i, str[i] - '0']++;
}
// Variable that stores the
// count of the numbers
int ans = 0;
// Traverse Input string and get the
// count of digits from count1 and
// count2 array such that difference
// b/w digit is 1 & store it int c1 &c2.
// And store it in variable c1 and c2
for(int i = 1; i < N - 1; i++)
{
if (str[i] == '9')
continue;
int c1 = count1[i - 1, str[i] - '0'];
int c2 = count2[i + 1, str[i] - '0' + 1];
if (c2 == 0)
continue;
// Update the ans
ans = (ans + (c1 * ((c2 * (c2 - 1) / 2))));
}
// Return the readonly count
return ans;
}// Driver codepublic static void Main(String[] args)
{ // Given String
String str = "224353";
int N = str.Length;
// Function call
Console.WriteLine(countStableNum(str, N));
}}// This code is contributed by Amit Katiyar |
Javascript
<script>// Javascript program for the above approach// Function to count the numbersfunction countStableNum(str, N)
{ // Array that stores the
// digits from left to right
var count1 = Array.from(Array(N), ()=>Array(10));
// Array that stores the
// digits from right to left
var count2 = Array.from(Array(N), ()=>Array(10));
// Initially both array store zero
for (var i = 0; i < N; i++)
for (var j = 0; j < 10; j++)
count1[i][j] = count2[i][j] = 0;
// Fill the table for count1 array
for (var i = 0; i < N; i++) {
if (i != 0) {
for (var j = 0; j < 10; j++) {
count1[i][j] += count1[i - 1][j];
}
}
// Update the count of current character
count1[i][str[i] - '0']++;
}
// Fill the table for count2 array
for (var i = N - 1; i >= 0; i--) {
if (i != N - 1) {
for (var j = 0; j < 10; j++) {
count2[i][j] += count2[i + 1][j];
}
}
// Update the count of current character
count2[i][str[i] - '0']++;
}
// Variable that stores the
// count of the numbers
var ans = 0;
// Traverse Input string and get the
// count of digits from count1 and
// count2 array such that difference
// b/w digit is 1 & store it var c1 &c2.
// And store it in variable c1 and c2
for (var i = 1; i < N - 1; i++) {
if (str[i] == '9')
continue;
var c1 = count1[i - 1][str[i] - '0'];
var c2 = count2[i + 1][str[i] - '0' + 1];
if (c2 == 0)
continue;
// Update the ans
ans = (ans
+ (c1 * ((c2 * (c2 - 1) / 2))));
}
// Return the final count
return ans;
}// Driver Code// Given Stringvar str = "224353";
var N = str.length;
// Function Calldocument.write( countStableNum(str, N));</script> |
Output:
1
Time Complexity: O(N)
Auxiliary Space Complexity: O(N)