Valid Anagram LeetCode Solution in Hindi (Hinglish)

Hindi mein samjho (Hinglish)

Neeche is topic ka Hinglish explanation hai, wahi jo hamare course lesson mein audio narration ke saath chalta hai. Code English mein hai, samjhana Hindi mein.

Valid Anagram chhota problem hai, par yahan ek reflex train hota hai jo baar baar use karoge: do collections ko unke contents se compare karna, order ignore karke. Count-then-cancel pattern yahan crisp ho gaya, toh Group Anagrams, Ransom Note, aur sliding-window anagram problems sab same problem feel karenge bas extra paint ke saath.

Problem ek line ka hai. Do strings s aur t diye gaye, return true agar t s ka anagram hai, warna false. Anagram ka matlab: t mein bilkul wahi letters hain jo s mein hain, exactly same frequencies ke saath, bas rearranged. Constraints kehte hain up to five times ten to the fourth characters aur sirf lowercase English letters, par follow-up explicitly Unicode ke baare mein poochta hai, isliye dono ke liye ready raho.

Pehle pattern pehchano. Jab bhi problem poochhe "kya yeh do collections mein same elements hain order ignore karke?": yeh hint hai frequencies count karne ka. Sorting ek tarika hai collection ko canonicalize karne ka. Frequency map usse fast tarika hai. Yahi pattern bahut disguises mein aata hai: group anagrams, ransom note, find all anagrams in a string, permutation in string.

Obvious solution se shuru karo: dono strings sort karo aur compare karo. Sorted versions identical hain to anagram hain. Python mein code ek line ka hai. Complexity O of n log n hai sort ke wajah se. Interview mein optimize karne se pehle yeh state karo: interviewer ko dikhta hai tumne simplest correct solution consider kiya cleverer cheez tak pohochne se pehle.

Ab asli unlock. Tumhe sorted order ki zaroorat hi nahi: tumhe sirf yeh jaanna hai har letter kitne baar aaya. Toh s ke har letter ko ek baar count karo, phir t walk karo aur har character pe count se ek letter cancel karo. Kabhi negative hone laga to t mein wo letter hai jo s mein nahi hai. Khatam hone par sab counts zero hain to anagram hain. Yeh ek pass har string pe, O of n.

Yeh hai optimal Python code. Gyaarah lines. Pehle length guard: different lengths means not an anagram, count karne ki zaroorat nahi. Phir s se ek frequency map build karo. Phir t walk karo: agar letter missing hai ya pehle se zero pe hai, immediately return false. Warna decrement karo aur continue. Loop complete hone par return true. Missing letter pe early exit hi adversarial inputs pe isse fast rakhta hai.

Chaar edge cases jo interviewers love karte hain. Different lengths: tumhara length guard ise constant time mein pakad leta hai. Dono empty: vacuously true, compare karne ko koi character hi nahi. Same string: har count one se wapas zero ho jaata hai, returns true. T mein ek extra character: tumhara missing-or-zero check us offending character pe false return karta hai, baaki scan kiye bina.

Time complexity O of n hai. Ek pass count build karne ka, ek pass cancel karne ka: total do n. Space complexity O of one hai jab lowercase English tak limited ho, kyunki count map mein at most chhabbis keys hain. Sort approach se compare karo jo O of n log n time hai aur O of one ya O of n space depending on language. n equals five times ten to the fourth ke liye, count approach lagbhag pachas hazaar operations mein chalta hai sort ke aath lakh ke against: clean dus guna ya zyada win.

India context. Valid Anagram TCS NQT, Infosys SP, aur Wipro Elite coding rounds mein aata hai, almost always plain form mein. Flipkart, Razorpay, PhonePe, aur Swiggy jaisi product companies isse harder string problems jaise Group Anagrams ya Find All Anagrams in a String ke pehle warm-up ki tarah use karti hain. Amazon aur Microsoft phone screens mein same paintaalis-minute slot mein Unicode aur grouping variants pe follow-up bhi pooch sakte hain.

Python solution

def is_anagram(s, t):
    if len(s) != len(t):
        return False
    count = {}
    for c in s:
        count[c] = count.get(c, 0) + 1
    for c in t:
        if c not in count or count[c] == 0:
            return False
        count[c] -= 1
    return True

Complexity: Sort and compare: time O(n log n), space O(n): sorted() returns a new list in Python; Sort and compare: time O(n log n), space O(n); Count then cancel (hash map): time O(n), space O(k): k = alphabet size.