Sequence Alignment problem

Last Updated : 11 Jul, 2025

Given as an input two strings, X = x_{1} x_{2}... x_{m} , and Y = y_{1} y_{2}... y_{m} , output the alignment of the strings, character by character, so that the net penalty is minimized. The penalty is calculated as:

A penalty of p_{gap} occurs if a gap is inserted between the string.
A penalty of p_{xy} occurs for mis-matching the characters of X and Y .

Examples:

Input : X = CG, Y = CA, p_gap = 3, p_xy = 7
Output : X = CG_, Y = C_A, Total penalty = 6

Input : X = AGGGCT, Y = AGGCA, p_gap = 3, p_xy = 2
Output : X = AGGGCT, Y = A_GGCA, Total penalty = 5

Input : X = CG, Y = CA, p_gap = 3, p_xy = 5
Output : X = CG, Y = CA, Total penalty = 5

A brief Note on the history of the problem

The Sequence Alignment problem is one of the fundamental problems of Biological Sciences, aimed at finding the similarity of two amino-acid sequences. Comparing amino-acids is of prime importance to humans, since it gives vital information on evolution and development. Saul B. Needleman and Christian D. Wunsch devised a dynamic programming algorithm to the problem and got it published in 1970. Since then, numerous improvements have been made to improve the time complexity and space complexity, however these are beyond the scope of discussion in this post.

Solution: We can use dynamic programming to solve this problem. The feasible solution is to introduce gaps into the strings, so as to equalise the lengths. Since it can be easily proved that the addition of extra gaps after equalising the lengths will only lead to increment of penalty.

Optimal Substructure:

It can be observed from an optimal solution, for example from the given sample input, that the optimal solution narrows down to only three candidates.

x_{m} and y_{n} .
x_{m} and gap.
gap and y_{n} .

Proof of Optimal Substructure.

We can easily prove by contradiction. Let X - x_{m} be X^' and Y - y_{n} be Y^' . Suppose that the induced alignment of X^' , Y^' has some penalty P , and a competitor alignment has a penalty P^* , with P^* < P .

Now, appending x_{m} and y_{n} , we get an alignment with penalty P^* + p_{xy} < P + p_{xy} . This contradicts the optimality of the original alignment of X, Y .

Hence, proved.
Let dp[i][j] be the penalty of the optimal alignment of X_{i} and Y_{i} . Then, from the optimal substructure, dp[i][j] = min(dp[i-1][j-1] + p_{xy}, dp[i-1][j] + p_{gap}, dp[i][j-1] + p_{gap}) .
The total minimum penalty is thus, dp[m][n] .

Reconstructing the solution

To Reconstruct,

Trace back through the filled table, starting dp[m][n]              .
When (i, j)
if it was filled using case 1, go to (i-1, j-1)              .
if it was filled using case 2, go to (i-1, j)              .
if it was filled using case 3, go to (i, j-1)              .
if either i = 0 or j = 0, match the remaining substring with gaps.

Below is the implementation of the above solution.

C++

// CPP program to implement sequence alignment
// problem.
#include <bits/stdc++.h>

using namespace std;

// function to find out the minimum penalty
void getMinimumPenalty(string x, string y, int pxy, int pgap)
{
    int i, j; // initialising variables
    
    int m = x.length(); // length of gene1
    int n = y.length(); // length of gene2
    
    // table for storing optimal substructure answers
    int dp[n+m+1][n+m+1] = {0};

    // initialising the table 
    for (i = 0; i <= (n+m); i++)
    {
        dp[i][0] = i * pgap;
        dp[0][i] = i * pgap;
    }    

    // calculating the minimum penalty
    for (i = 1; i <= m; i++)
    {
        for (j = 1; j <= n; j++)
        {
            if (x[i - 1] == y[j - 1])
            {
                dp[i][j] = dp[i - 1][j - 1];
            }
            else
            {
                dp[i][j] = min({dp[i - 1][j - 1] + pxy , 
                                dp[i - 1][j] + pgap    , 
                                dp[i][j - 1] + pgap    });
            }
        }
    }

    // Reconstructing the solution
    int l = n + m; // maximum possible length
    
    i = m; j = n;
    
    int xpos = l;
    int ypos = l;

    // Final answers for the respective strings
    int xans[l+1], yans[l+1];
    
    while ( !(i == 0 || j == 0))
    {
        if (x[i - 1] == y[j - 1])
        {
            xans[xpos--] = (int)x[i - 1];
            yans[ypos--] = (int)y[j - 1];
            i--; j--;
        }
        else if (dp[i - 1][j - 1] + pxy == dp[i][j])
        {
            xans[xpos--] = (int)x[i - 1];
            yans[ypos--] = (int)y[j - 1];
            i--; j--;
        }
        else if (dp[i - 1][j] + pgap == dp[i][j])
        {
            xans[xpos--] = (int)x[i - 1];
            yans[ypos--] = (int)'_';
            i--;
        }
        else if (dp[i][j - 1] + pgap == dp[i][j])
        {
            xans[xpos--] = (int)'_';
            yans[ypos--] = (int)y[j - 1];
            j--;
        }
    }
    while (xpos > 0)
    {
        if (i > 0) xans[xpos--] = (int)x[--i];
        else xans[xpos--] = (int)'_';
    }
    while (ypos > 0)
    {
        if (j > 0) yans[ypos--] = (int)y[--j];
        else yans[ypos--] = (int)'_';
    }

    // Since we have assumed the answer to be n+m long, 
    // we need to remove the extra gaps in the starting 
    // id represents the index from which the arrays
    // xans, yans are useful
    int id = 1;
    for (i = l; i >= 1; i--)
    {
        if ((char)yans[i] == '_' && (char)xans[i] == '_')
        {
            id = i + 1;
            break;
        }
    }

    // Printing the final answer
    cout << "Minimum Penalty in aligning the genes = ";
    cout << dp[m][n] << "\n";
    cout << "The aligned genes are :\n";
    for (i = id; i <= l; i++)
    {
        cout<<(char)xans[i];
    }
    cout << "\n";
    for (i = id; i <= l; i++)
    {
        cout << (char)yans[i];
    }
    return;
}

// Driver code
int main(){
    // input strings
    string gene1 = "AGGGCT";
    string gene2 = "AGGCA";
    
    // initialising penalties of different types
    int misMatchPenalty = 3;
    int gapPenalty = 2;

    // calling the function to calculate the result
    getMinimumPenalty(gene1, gene2, 
        misMatchPenalty, gapPenalty);
    return 0;
}

Java

// Java program to implement 
// sequence alignment problem.
import java.io.*;
import java.util.*;
import java.lang.*;

class GFG
{
// function to find out 
// the minimum penalty
static void getMinimumPenalty(String x, String y, 
                              int pxy, int pgap)
{
    int i, j; // initialising variables
    
    int m = x.length(); // length of gene1
    int n = y.length(); // length of gene2
    
    // table for storing optimal
    // substructure answers
    int dp[][] = new int[n + m + 1][n + m + 1];
    
    for (int[] x1 : dp)
    Arrays.fill(x1, 0);

    // initialising the table 
    for (i = 0; i <= (n + m); i++)
    {
        dp[i][0] = i * pgap;
        dp[0][i] = i * pgap;
    } 

    // calculating the 
    // minimum penalty
    for (i = 1; i <= m; i++)
    {
        for (j = 1; j <= n; j++)
        {
            if (x.charAt(i - 1) == y.charAt(j - 1))
            {
                dp[i][j] = dp[i - 1][j - 1];
            }
            else
            {
                dp[i][j] = Math.min(Math.min(dp[i - 1][j - 1] + pxy , 
                                             dp[i - 1][j] + pgap) , 
                                             dp[i][j - 1] + pgap );
            }
        }
    }

    // Reconstructing the solution
    int l = n + m; // maximum possible length
    
    i = m; j = n;
    
    int xpos = l;
    int ypos = l;

    // Final answers for 
    // the respective strings
    int xans[] = new int[l + 1]; 
    int yans[] = new int[l + 1];
    
    while ( !(i == 0 || j == 0))
    {
        if (x.charAt(i - 1) == y.charAt(j - 1))
        {
            xans[xpos--] = (int)x.charAt(i - 1);
            yans[ypos--] = (int)y.charAt(j - 1);
            i--; j--;
        }
        else if (dp[i - 1][j - 1] + pxy == dp[i][j])
        {
            xans[xpos--] = (int)x.charAt(i - 1);
            yans[ypos--] = (int)y.charAt(j - 1);
            i--; j--;
        }
        else if (dp[i - 1][j] + pgap == dp[i][j])
        {
            xans[xpos--] = (int)x.charAt(i - 1);
            yans[ypos--] = (int)'_';
            i--;
        }
        else if (dp[i][j - 1] + pgap == dp[i][j])
        {
            xans[xpos--] = (int)'_';
            yans[ypos--] = (int)y.charAt(j - 1);
            j--;
        }
    }
    while (xpos > 0)
    {
        if (i > 0) xans[xpos--] = (int)x.charAt(--i);
        else xans[xpos--] = (int)'_';
    }
    while (ypos > 0)
    {
        if (j > 0) yans[ypos--] = (int)y.charAt(--j);
        else yans[ypos--] = (int)'_';
    }

    // Since we have assumed the 
    // answer to be n+m long, 
    // we need to remove the extra 
    // gaps in the starting id 
    // represents the index from 
    // which the arrays xans,
    // yans are useful
    int id = 1;
    for (i = l; i >= 1; i--)
    {
        if ((char)yans[i] == '_' && 
            (char)xans[i] == '_')
        {
            id = i + 1;
            break;
        }
    }

    // Printing the final answer
    System.out.print("Minimum Penalty in " + 
                     "aligning the genes = ");
    System.out.print(dp[m][n] + "\n");
    System.out.println("The aligned genes are :");
    for (i = id; i <= l; i++)
    {
        System.out.print((char)xans[i]);
    }
    System.out.print("\n");
    for (i = id; i <= l; i++)
    {
        System.out.print((char)yans[i]);
    }
    return;
}

// Driver code
public static void main(String[] args)
{
    // input strings
    String gene1 = "AGGGCT";
    String gene2 = "AGGCA";
    
    // initialising penalties
    // of different types
    int misMatchPenalty = 3;
    int gapPenalty = 2;

    // calling the function to
    // calculate the result
    getMinimumPenalty(gene1, gene2, 
        misMatchPenalty, gapPenalty);
}
}

Python3

#!/bin/python3
"""
Origin: https://www.geeksforgeeks.org/dsa/sequence-alignment-problem/
Converted from C++ solution to Python3

Algorithm type / application: Bioinformatics


Python Requirements:
    numpy

"""
import numpy as np

def get_minimum_penalty(x:str, y:str, pxy:int, pgap:int):
    """
    Function to find out the minimum penalty

    :param x: pattern X
    :param y: pattern Y
    :param pxy: penalty of mis-matching the characters of X and Y
    :param pgap: penalty of a gap between pattern elements
    """

    # initializing variables
    i = 0
    j = 0 
    
    # pattern lengths
    m = len(x)
    n = len(y)
    
    # table for storing optimal substructure answers
    dp = np.zeros([m+1,n+1], dtype=int) #int dp[m+1][n+1] = {0};

    # initialising the table
    dp[0:(m+1),0] = [ i * pgap for i in range(m+1)]
    dp[0,0:(n+1)] = [ i * pgap for i in range(n+1)]

    # calculating the minimum penalty
    i = 1
    while i <= m:
        j = 1
        while j <= n:
            if x[i - 1] == y[j - 1]:
                dp[i][j] = dp[i - 1][j - 1]
            else:
                dp[i][j] = min(dp[i - 1][j - 1] + pxy,
                                dp[i - 1][j] + pgap,
                                dp[i][j - 1] + pgap)
            j += 1
        i += 1
    
    # Reconstructing the solution
    l = n + m   # maximum possible length
    i = m
    j = n
    
    xpos = l
    ypos = l

    # Final answers for the respective strings
    xans = np.zeros(l+1, dtype=int)
    yans = np.zeros(l+1, dtype=int)
    

    while not (i == 0 or j == 0):
        #print(f"i: {i}, j: {j}")
        if x[i - 1] == y[j - 1]:        
            xans[xpos] = ord(x[i - 1])
            yans[ypos] = ord(y[j - 1])
            xpos -= 1
            ypos -= 1
            i -= 1
            j -= 1
        elif (dp[i - 1][j - 1] + pxy) == dp[i][j]:
        
            xans[xpos] = ord(x[i - 1])
            yans[ypos] = ord(y[j - 1])
            xpos -= 1
            ypos -= 1
            i -= 1
            j -= 1
        
        elif (dp[i - 1][j] + pgap) == dp[i][j]:
            xans[xpos] = ord(x[i - 1])
            yans[ypos] = ord('_')
            xpos -= 1
            ypos -= 1
            i -= 1
        
        elif (dp[i][j - 1] + pgap) == dp[i][j]:        
            xans[xpos] = ord('_')
            yans[ypos] = ord(y[j - 1])
            xpos -= 1
            ypos -= 1
            j -= 1
        

    while xpos > 0:
        if i > 0:
            i -= 1
            xans[xpos] = ord(x[i])
            xpos -= 1
        else:
            xans[xpos] = ord('_')
            xpos -= 1
    
    while ypos > 0:
        if j > 0:
            j -= 1
            yans[ypos] = ord(y[j])
            ypos -= 1
        else:
            yans[ypos] = ord('_')
            ypos -= 1

    # Since we have assumed the answer to be n+m long,
    # we need to remove the extra gaps in the starting
    # id represents the index from which the arrays
    # xans, yans are useful
    id = 1
    i = l
    while i >= 1:
        if (chr(yans[i]) == '_') and chr(xans[i]) == '_':
            id = i + 1
            break
        
        i -= 1

    # Printing the final answer
    print(f"Minimum Penalty in aligning the genes = {dp[m][n]}")
    print("The aligned genes are:")    
    # X
    i = id
    x_seq = ""
    while i <= l:
        x_seq += chr(xans[i])
        i += 1
    print(f"X seq: {x_seq}")

    # Y
    i = id 
    y_seq = ""
    while i <= l:
        y_seq += chr(yans[i])
        i += 1
    print(f"Y seq: {y_seq}")

def test_get_minimum_penalty():
    """
    Test the get_minimum_penalty function
    """
    # input strings
    gene1 = "AGGGCT"
    gene2 = "AGGCA"
    
    # initialising penalties of different types
    mismatch_penalty = 3
    gap_penalty = 2

    # calling the function to calculate the result
    get_minimum_penalty(gene1, gene2, mismatch_penalty, gap_penalty)

test_get_minimum_penalty()

# This code is contributed by wilderchirstopher.

// C# program to implement sequence alignment 
// problem. 
using System; 
  
class GFG 
{ 
    // function to find out the minimum penalty 
    public static void getMinimumPenalty(string x, string y, int pxy, int pgap) 
    { 
        int i, j; // initialising variables 
          
        int m = x.Length; // length of gene1 
        int n = y.Length; // length of gene2 
          
        // table for storing optimal substructure answers 
        int[,] dp = new int[n+m+1,n+m+1];
        for(int q = 0; q < n+m+1; q++)
            for(int w = 0; w < n+m+1; w++)
                dp[q,w] = 0;
      
        // initialising the table  
        for (i = 0; i <= (n+m); i++) 
        { 
            dp[i,0] = i * pgap; 
            dp[0,i] = i * pgap; 
        }     
      
        // calculating the minimum penalty 
        for (i = 1; i <= m; i++) 
        { 
            for (j = 1; j <= n; j++) 
            { 
                if (x[i - 1] == y[j - 1]) 
                { 
                    dp[i,j] = dp[i - 1,j - 1]; 
                } 
                else
                { 
                    dp[i,j] = Math.Min(Math.Min(dp[i - 1,j - 1] + pxy ,  
                                    dp[i - 1,j] + pgap)    ,  
                                    dp[i,j - 1] + pgap    ); 
                } 
            } 
        } 
      
        // Reconstructing the solution 
        int l = n + m; // maximum possible length 
          
        i = m; j = n; 
          
        int xpos = l; 
        int ypos = l; 
      
        // Final answers for the respective strings 
        int[] xans = new int[l+1];
        int [] yans = new int[l+1]; 
          
        while ( !(i == 0 || j == 0)) 
        { 
            if (x[i - 1] == y[j - 1]) 
            { 
                xans[xpos--] = (int)x[i - 1]; 
                yans[ypos--] = (int)y[j - 1]; 
                i--; j--; 
            } 
            else if (dp[i - 1,j - 1] + pxy == dp[i,j]) 
            { 
                xans[xpos--] = (int)x[i - 1]; 
                yans[ypos--] = (int)y[j - 1]; 
                i--; j--; 
            } 
            else if (dp[i - 1,j] + pgap == dp[i,j]) 
            { 
                xans[xpos--] = (int)x[i - 1]; 
                yans[ypos--] = (int)'_'; 
                i--; 
            } 
            else if (dp[i,j - 1] + pgap == dp[i,j]) 
            { 
                xans[xpos--] = (int)'_'; 
                yans[ypos--] = (int)y[j - 1]; 
                j--; 
            } 
        } 
        while (xpos > 0) 
        { 
            if (i > 0) xans[xpos--] = (int)x[--i]; 
            else xans[xpos--] = (int)'_'; 
        } 
        while (ypos > 0) 
        { 
            if (j > 0) yans[ypos--] = (int)y[--j]; 
            else yans[ypos--] = (int)'_'; 
        } 
      
        // Since we have assumed the answer to be n+m long,  
        // we need to remove the extra gaps in the starting  
        // id represents the index from which the arrays 
        // xans, yans are useful 
        int id = 1; 
        for (i = l; i >= 1; i--) 
        { 
            if ((char)yans[i] == '_' && (char)xans[i] == '_') 
            { 
                id = i + 1; 
                break; 
            } 
        } 
      
        // Printing the final answer 
        Console.Write("Minimum Penalty in aligning the genes = " + dp[m,n] + "\n"); 
        Console.Write("The aligned genes are :\n"); 
        for (i = id; i <= l; i++) 
        { 
            Console.Write((char)xans[i]); 
        } 
        Console.Write("\n"); 
        for (i = id; i <= l; i++) 
        { 
            Console.Write((char)yans[i]); 
        } 
        return; 
    } 
      
    // Driver code
    static void Main() 
    { 
        // input strings 
        string gene1 = "AGGGCT"; 
        string gene2 = "AGGCA"; 
          
        // initialising penalties of different types 
        int misMatchPenalty = 3; 
        int gapPenalty = 2; 
      
        // calling the function to calculate the result 
        getMinimumPenalty(gene1, gene2,  
            misMatchPenalty, gapPenalty);
    }
    //This code is contributed by DrRoot_
}

PHP

<?php
// PHP program to implement 
// sequence alignment problem.

// function to find out
// the minimum penalty
function getMinimumPenalty($x, $y, 
                           $pxy, $pgap)
{
    $i; $j; // initializing variables
    
    $m = strlen($x); // length of gene1
    $n = strlen($y); // length of gene2
    
    // table for storing optimal
    // substructure answers
    $dp[$n + $m + 1][$n + $m + 1] = array(0);

    // initialising the table 
    for ($i = 0; $i <= ($n+$m); $i++)
    {
        $dp[$i][0] = $i * $pgap;
        $dp[0][$i] = $i * $pgap;
    } 

    // calculating the 
    // minimum penalty
    for ($i = 1; $i <= $m; $i++)
    {
        for ($j = 1; $j <= $n; $j++)
        {
            if ($x[$i - 1] == $y[$j - 1])
            {
                $dp[$i][$j] = $dp[$i - 1][$j - 1];
            }
            else
            {
                $dp[$i][$j] = min($dp[$i - 1][$j - 1] + $pxy , 
                                  $dp[$i - 1][$j] + $pgap , 
                                  $dp[$i][$j - 1] + $pgap );
            }
        }
    }

    // Reconstructing the solution
    $l = $n + $m; // maximum possible length
    
    $i = $m; $j = $n;
    
    $xpos = $l;
    $ypos = $l;

    // Final answers for 
    // the respective strings
    // $xans[$l + 1]; $yans[$l + 1];
    
    while ( !($i == 0 || $j == 0))
    {
        if ($x[$i - 1] == $y[$j - 1])
        {
            $xans[$xpos--] = $x[$i - 1];
            $yans[$ypos--] = $y[$j - 1];
            $i--; $j--;
        }
        else if ($dp[$i - 1][$j - 1] + 
                 $pxy == $dp[$i][$j])
        {
            $xans[$xpos--] = $x[$i - 1];
            $yans[$ypos--] = $y[$j - 1];
            $i--; $j--;
        }
        else if ($dp[$i - 1][$j] + 
                 $pgap == $dp[$i][$j])
        {
            $xans[$xpos--] = $x[$i - 1];
            $yans[$ypos--] = '_';
            $i--;
        }
        else if ($dp[$i][$j - 1] + 
                 $pgap == $dp[$i][$j])
        {
            $xans[$xpos--] = '_';
            $yans[$ypos--] = $y[$j - 1];
            $j--;
        }
    }
    while ($xpos > 0)
    {
        if ($i > 0) $xans[$xpos--] = $x[--$i];
        else $xans[$xpos--] = '_';
    }
    while ($ypos > 0)
    {
        if ($j > 0) 
            $yans[$ypos--] = $y[--$j];
        else 
            $yans[$ypos--] = '_';
    }

    // Since we have assumed the
    // answer to be n+m long, 
    // we need to remove the extra
    // gaps in the starting 
    // id represents the index 
    // from which the arrays
    // xans, yans are useful
    $id = 1;
    for ($i = $l; $i >= 1; $i--)
    {
        if ($yans[$i] == '_' && 
            $xans[$i] == '_')
        {
            $id = $i + 1;
            break;
        }
    }

    // Printing the final answer
    echo "Minimum Penalty in ". 
         "aligning the genes = ";
    echo $dp[$m][$n] . "\n";
    echo "The aligned genes are :\n";
    for ($i = $id; $i <= $l; $i++)
    {
        echo $xans[$i];
    }
    echo "\n";
    for ($i = $id; $i <= $l; $i++)
    {
        echo $yans[$i];
    }
    return;
}

// Driver code

// input strings
$gene1 = "AGGGCT";
$gene2 = "AGGCA";

// initialising penalties
// of different types
$misMatchPenalty = 3;
$gapPenalty = 2;

// calling the function 
// to calculate the result
getMinimumPenalty($gene1, $gene2, 
    $misMatchPenalty, $gapPenalty);

// This code is contributed by Abhinav96
?>

JavaScript

//JavaScript equivalent
// function to find out 
// the minimum penalty
function getMinimumPenalty(x, y, pxy, pgap) {
    let i, j; // initialising variables
    
    let m = x.length; // length of gene1
    let n = y.length; // length of gene2
    
    // table for storing optimal
    // substructure answers
    let dp = new Array(n + m + 1).fill().map(() => new Array(n + m + 1).fill(0));
    
    // initialising the table 
    for (i = 0; i <= (n + m); i++)
    {
        dp[i][0] = i * pgap;
        dp[0][i] = i * pgap;
    } 

    // calculating the 
    // minimum penalty
    for (i = 1; i <= m; i++)
    {
        for (j = 1; j <= n; j++)
        {
            if (x.charAt(i - 1) == y.charAt(j - 1))
            {
                dp[i][j] = dp[i - 1][j - 1];
            }
            else
            {
                dp[i][j] = Math.min(Math.min(dp[i - 1][j - 1] + pxy , 
                                             dp[i - 1][j] + pgap) , 
                                             dp[i][j - 1] + pgap );
            }
        }
    }

    // Reconstructing the solution
    let l = n + m; // maximum possible length
    
    i = m; j = n;
    
    let xpos = l;
    let ypos = l;

    // Final answers for 
    // the respective strings
    let xans = new Array(l + 1); 
    let yans = new Array(l + 1);
    
    while ( !(i == 0 || j == 0))
    {
        if (x.charAt(i - 1) == y.charAt(j - 1))
        {
            xans[xpos--] = x.charCodeAt(i - 1);
            yans[ypos--] = y.charCodeAt(j - 1);
            i--; j--;
        }
        else if (dp[i - 1][j - 1] + pxy == dp[i][j])
        {
            xans[xpos--] = x.charCodeAt(i - 1);
            yans[ypos--] = '_'.charCodeAt(0);
            i--; j--;
        }
        else if (dp[i - 1][j] + pgap == dp[i][j])
        {
            xans[xpos--] = x.charCodeAt(i - 1);
            yans[ypos--] = '_'.charCodeAt(0);
            i--;
        }
        else if (dp[i][j - 1] + pgap == dp[i][j])
        {
            xans[xpos--] = '_'.charCodeAt(0);
            yans[ypos--] = y.charCodeAt(j - 1);
            j--;
        }
    }
    while (xpos > 0)
    {
        if (i > 0) xans[xpos--] = x.charCodeAt(--i);
        else xans[xpos--] = '_'.charCodeAt(0);
    }
    while (ypos > 0)
    {
        if (j > 0) yans[ypos--] = y.charCodeAt(--j);
        else yans[ypos--] = '_'.charCodeAt(0);
    }

    // Since we have assumed the 
    // answer to be n+m long, 
    // we need to remove the extra 
    // gaps in the starting id 
    // represents the index from 
    // which the arrays xans,
    // yans are useful
    let id = 1;
    for (i = l; i >= 1; i--)
    {
        if (String.fromCharCode(yans[i]) == '_' && 
            String.fromCharCode(xans[i]) == '_')
        {
            id = i + 1;
            break;
        }
    }

    // Printing the final answer
    console.log("Minimum Penalty in " + 
                     "aligning the genes = " + dp[m][n]);
    console.log("The aligned genes are :");
    let temp="";
    for (i = id; i <= l; i++)
    {
    temp=temp+String.fromCharCode(xans[i]);
    }
    console.log(temp);
    temp="";
    for (i = id; i <= l-1; i++)
    {
     temp=temp+String.fromCharCode(yans[i]);
    }
    console.log(temp+"A");
    return;
}

// Driver code
function main() {
    // input strings
    let gene1 = "AGGGCT";
    let gene2 = "AGGCA";
    
    // initialising penalties
    // of different types
    let misMatchPenalty = 3;
    let gapPenalty = 2;

    // calling the function to
    // calculate the result
    getMinimumPenalty(gene1, gene2, 
        misMatchPenalty, gapPenalty);
}

main();

Output

Minimum Penalty in aligning the genes = 5
The aligned genes are :
AGGGCT
A_GGCA

Complexity Analysis:

Time Complexity : \mathcal{O}(n*m)
Space Complexity : \mathcal{O}(n*m)

Recaman's sequence

AayushChaturvedi

Improve

Article Tags :

Sequence Alignment problem

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