Subarrays with distinct elements

#1

“subarrays with distinct elements” ki approach bhi bta do?

#2

hello @CODER_JATIN

pehle tum ye batao , agar tumhe L lenght ki array di hui hai to uske smaller subarray length ka sum kitna banega (starting from one end)?

#3

mere hisab se n*(n+1)/2 banne chahiye

#4

ha . . . . . .

to isproblem me bhi yahi use karna hai, do pointer rakho(ek pointer starting point ke liye aur ek point ending point ke liye)
first element se expand karna suru karo , aur tabtak expand karo jabtak distinct element aa rahe hai. maan lo total L expand hua,
to isko add karlo apne answer me.
ans+=(L * (L+1)/2)

ab kyunki window me aur dintinct elements nahi aa sakte hai, hum starting elemnt ko remove karneg aur window shrink karnege.

isiko ko repeat karna hai baaki remaining element pe

#5

@CODER_JATIN

ab isko dekho aur batao kuch samjh aa raha hai ya nahi.

int sumOflength(int *arr, int n) {
    unordered_set<int> s;
    int j=0, ans=0;
    for(int i=0;i<n;i++) {  // j hamara right pointer hai , i hamara left pointer hai
        while(j<n and s.find(arr[j])==s.end()) { // expand kar rahe hai jabtak distint element hai
            s.insert(arr[j]);
            j++;
        }
        ans+= ((j-i)*(j-i+1))/2;// j-i jo hai wo L hai 
        s.erase(arr[i]);//starting element ko erase kar rahe hai kyuni ab window me distinct element nahi hai
    }
    return ans;
}
#6

bhaiya ye pehla point to smjh aa gya window kb tk expand krni hai or kab shrink karni hai

#7

ab dusra point smjh rha hu . . . . .

#8

ans+= ((j-i)*(j-i+1))/2;,
bhaiya isse to no of subarrays niklenge na us length mai kitne subarrays hai vo pta lgega na isse to?

#9

ha … . . . .
ab kyunki jo subarray hamne nikali hai usme saare element distinct hai .
to is subarray be jitne bhi possible smaller subarray(starting from i) honge wo sab bhi distinct element hold karnege haina?

isilye wo formula use kar rahe hai.

#10

bhaiya maanlo i=0 hai or j=4 hai ,
to jab ham ye formula use karenge tb (0 to 4) length mai jitne bhi subarrays hai vo aa gye hmare paas , ab jab ham loop ki next iteration m aayenge tb , i=1 hojega or j=4 hi rhega to jo ab firse ham subarrays nikaalenge, ab jo subarrays aaye vo to ham pehle hi i=0 waali iteration mai nikaal chuke hai already?
to repeat ni hore subarrrays?

#11

bhaiya maanlo i=0 hai or j=4 hai ,
to jab ham ye formula use karenge tb (0 to 4) length mai jitne bhi subarrays hai vo aa gye hmare paas , ab jab ham loop ki next iteration m aayenge tb , i=1 hojega or j=4 hi rhega to jo ab firse ham subarrays nikaalenge, ab jo subarrays aaye vo to ham pehle hi i=0 waali iteration mai nikaal chuke hai already?
to repeat ni hore subarrrays?

#12

bro ,sum of length hai na sawal me.

jaise tume i=0 aur j=4 hai to 4-0 = 4 hogya ?
[0] , [0.1] , [0,1,2] , [0,1,2,3]
inka lenght ka sum karo?
1 + 2 + 3 + 4 -> 10 ( jo ki 4 * (4+1)/2 ke barabar hai)

maine subarray word galat use kar liya isiliye tumhe confusion ho gayi sayad

#13

hanji bhaiya , mai vahi to confuse ho gya

#14

hanji bhaiya , mai vahi to confuse ho gya . .

#15

abhi bhi hai kya confusion… . . .

L length ki subarray me smaller kitni length hongi
1 (1 lengt ki ek hogi )
2 (2 length ki ek hogi)
3 (3 length ki ek hogi)
L (L length ki ek hogi)

in length ka sum kita hoga
1 + 2 + 3 + 4 + …L -> (L *(L+1) /2)

ab bhi confusion hai to batao

#16

bhaiya je jo formula hai (l)*(l+1)/2 isse hamey no of subarrays pata lagte hai ek particular length mai,
ham saare subarrays ko plus kar rhe to hamey subarrays ki length pta lg rhi hai.

#17

ss

is dekho
ab ye 4 lenght ki array hai .
to ab un sab subarray ka lenght ka sum batao jisme 0 ho.
to answer kya hoga tumhara?

#18

10 hoga? . . . . . . …

#19

ha…
[0] -> 1 length
[0,1] -> 2 length
[0,1,2] -> 3 length
[0,1,2,3] -> 4 length
1+2 + 3 + 4 =10 ( using formula 4*(4+1)/2 = 10 )
yahi bata raha tha tabse , wo suru me swal mujhe thoda clear kar dena chaiye tha , confusion avoid karne ke lie.

#20

kuch confusion ho to batao