⚽ Live Match: dsa/check whether two strings contain same characters in same order

Given two strings s1 and s2, find whether the two strings contain the same characters that occur in the same order. For example string "Geeks" and string "Geks" contain the same characters in same order.

Examples:

Input: s1 = "Geeks", s2 = "Geks"
Output: true
Explanation: Both strings follow the same character order: G, e, k, s. The first string has one extra repeated e, but the sequence of characters is still the same.

Input: s1 = "Arnab", s2 = "Andrew"
Output: false
Explanation: The character order is not the same in both strings, so the strings do not match.

Try It Yourself

Table of Content

[Naive Approach] Build Distinct Character Sequences - O(n + m) Time and O(n + m) Space
[Expected Approach] Two Pointer Traversal - O(n + m) Time and O(1) Space

[Naive Approach] Build Distinct Character Sequences - O(n + m) Time and O(n + m) Space

The idea is to build a new sequence for each string by removing consecutive duplicate characters. After processing both strings, compare the resulting sequences. If they are identical, then both strings contain the same characters in the same order.

C++

#include <iostream>
#include <string>
using namespace std;

bool sameSeq(string &s1, string &s2)
{

    string seq1, seq2;

    // Build sequence for first string
    for (char ch : s1)
    {
        if (seq1.empty() || seq1.back() != ch)
            seq1 += ch;
    }

    // Build sequence for second string
    for (char ch : s2)
    {
        if (seq2.empty() || seq2.back() != ch)
            seq2 += ch;
    }

    return seq1 == seq2;
}

int main()
{
    string s1 = "Geeks";
    string s2 = "Geks";

    if (sameSeq(s1, s2))
        cout << "true";
    else
        cout << "false";

    return 0;
}

Java

public class GFG {
    public static boolean sameSeq(String s1, String s2)
    {
        StringBuilder seq1 = new StringBuilder(),
                      seq2 = new StringBuilder();

        // Build sequence for first string
        for (char ch : s1.toCharArray()) {
            if (seq1.length() == 0
                || seq1.charAt(seq1.length() - 1) != ch)
                seq1.append(ch);
        }

        // Build sequence for second string
        for (char ch : s2.toCharArray()) {
            if (seq2.length() == 0
                || seq2.charAt(seq2.length() - 1) != ch)
                seq2.append(ch);
        }

        return seq1.toString().equals(seq2.toString());
    }

    public static void main(String[] args)
    {
        String s1 = "Geeks";
        String s2 = "Geks";

        if (sameSeq(s1, s2))
            System.out.println("true");
        else
            System.out.println("false");
    }
}

Python

def sameSeq(s1, s2):

    seq1, seq2 = '', ''

    # Build sequence for first string
    for ch in s1:
        if not seq1 or seq1[-1] != ch:
            seq1 += ch

    # Build sequence for second string
    for ch in s2:
        if not seq2 or seq2[-1] != ch:
            seq2 += ch

    return seq1 == seq2

if __name__ == "__main__":
    s1 = "Geeks"
    s2 = "Geks"

    if sameSeq(s1, s2):
        print("true")
    else:
        print("false")

using System;

class GFG {
    static bool sameSeq(string s1, string s2)
    {
        string seq1 = "", seq2 = "";

        // Build sequence for first string
        foreach(char ch in s1)
        {
            if (seq1.Length == 0
                || seq1[seq1.Length - 1] != ch)
                seq1 += ch;
        }

        // Build sequence for second string
        foreach(char ch in s2)
        {
            if (seq2.Length == 0
                || seq2[seq2.Length - 1] != ch)
                seq2 += ch;
        }

        return seq1 == seq2;
    }

    static void Main()
    {
        string s1 = "Geeks";
        string s2 = "Geks";

        if (sameSeq(s1, s2))
            Console.Write("true");
        else
            Console.Write("false");
    }
}

JavaScript

function sameSeq(s1, s2)
{
    let seq1 = "", seq2 = "";

    // Build sequence for first string
    for (let ch of s1) {
        if (seq1.length === 0
            || seq1[seq1.length - 1] !== ch)
            seq1 += ch;
    }

    // Build sequence for second string
    for (let ch of s2) {
        if (seq2.length === 0
            || seq2[seq2.length - 1] !== ch)
            seq2 += ch;
    }

    return seq1 === seq2;
}

// Driver code
let s1 = "Geeks";
let s2 = "Geks";

if (sameSeq(s1, s2))
    console.log("true");
else
    console.log("false");

Output

true

[Expected Approach] Two Pointer Traversal - O(n + m) Time and O(1) Space

The idea is to use two pointers to traverse both strings simultaneously. Compare the current characters of both strings and, if they match, skip all their consecutive occurrences. If a mismatch is found, return false. If both strings are completely traversed together, then they contain the same characters in the same order.

Let us understand with example:
Input: s1 = "Geeks", s2 = "Geks"

Initialize i = 0 and j = 0. Characters 'G' and 'G' match, so skip their occurrences. Now i = 1, j = 1.
Characters 'e' and 'e' match. Skip both 'e' characters in s1 and the single 'e' in s2. Now i = 3, j = 2.
Characters 'k' and 'k' match, so skip them. Now i = 4, j = 3.
Characters 's' and 's' match, so skip them. Now i = 5, j = 4.
Both strings are completely traversed, so return true.

C++

#include <iostream>
#include <string>
using namespace std;

bool sameSeq(string &s1, string &s2)
{
    int i = 0, j = 0;
    int n = s1.size(), m = s2.size();

    while (i < n && j < m)
    {

        // Current character groups must match
        if (s1[i] != s2[j])
            return false;

        char c1 = s1[i];
        char c2 = s2[j];

        // Skip all consecutive occurrences
        while (i < n && s1[i] == c1)
            i++;

        while (j < m && s2[j] == c2)
            j++;
    }

    // Both strings must be fully processed
    return (i == n && j == m);
}

// Driver code
int main()
{
    string s1 = "Geeks";
    string s2 = "Geks";

    if (sameSeq(s1, s2))
        cout << "true";
    else
        cout << "false";

    return 0;
}