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Tcs coding 1.D 2.break 3.arrayindexout 4.B 5.B 6.D 7.change control 8-37 9-10 10-2 Telegram - https://t.me/Coding_solution_0

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Tcs coding 1.D 2.break 3.arrayindexout 4.B 5.B 6.D 7.change control 8.37 https://t.me/Coding_solution_0 9.10

Tcs coding 1.D 2.break 3.arrayindexout 4.B 5.B 6.D 7.change control 8.37 Telegram - https://t.me/Coding_solution_0

Tcs coding 1.D 2.break 3.arrayindexout 4.B 5.B 6.D 7.change control Telegram -https://t.me/Coding_solution_0

Tcs coding 1.D 2.break 3.arrayindexout 4.B 5.B 6.D Telegram - https://t.me/Coding_solution_0

Reasoning: 1.)655 2.)1814 3.)700 4.)266 5.)143 6.)368 7.)WSEF 8.)UDED 9.)ACIP 10.)2,4,1,3 11.)4,5,3,1,2,6 12.)4,5,1,2,3 13.)CDEFJ 14.)OFP 15.)Brother in law 16.)W 17.)The date in Both statement 1 and 2 together not sufficient 18.)in front of c 19.) second right of S 20.)B 21.)D 22.)Eiether conclusion I or II 23.)Only statement 1 & 3 are implicit 24.)only argument 1 is implicit 25.)Only course of Action II follows 26.)Both A and R true but R is not correct explaination of A 27.)C 28.)B 29.)C 30.)D Paid coding Solution Available 😜 All test case pass @Wren_0 @Wren_0

class Solution: def calculateMinimumHP(self, dungeon: List[List[int]]) -> int: def oneIfNotPos(a): return a if a > 0 else 1 def f(i: int, j: int) -> int: cached = self.cache.get((i, j)) if cached: return cached answer = min(oneIfNotPos(f(i, j + 1) - dungeon[i][j]), \ oneIfNotPos(f(i + 1, j) - dungeon[i][j])) self.cache[(i, j)] = answer return answer m = len(dungeon) n = len(dungeon[0]) # init cache to remove ifs from f self.cache = {(m - 1, n - 1): oneIfNotPos(1 - dungeon[m - 1][n - 1])} for i in range(0, m): self.cache[(i, n)] = 1000 000000 # inf for j in range(0, n): self.cache[(m, j)] = 10000 00000 # inf return f(0, 0) Python Dungeon Hunters Telegram -https://t.me/Coding_solution_0

class Solution: def calculateMinimumHP(self, dungeon: List[List[int]]) -> int: def oneIfNotPos(a): return a if a > 0 else 1 def f(i: int, j: int) -> int: cached = self.cache.get((i, j)) if cached: return cached answer = min(oneIfNotPos(f(i, j + 1) - dungeon[i][j]), \ oneIfNotPos(f(i + 1, j) - dungeon[i][j])) self.cache[(i, j)] = answer return answer m = len(dungeon) n = len(dungeon[0]) # init cache to remove ifs from f self.cache = {(m - 1, n - 1): oneIfNotPos(1 - dungeon[m - 1][n - 1])} for i in range(0, m): self.cache[(i, n)] = 1000 000000 # inf for j in range(0, n): self.cache[(m, j)] = 10000 00000 # inf return f(0, 0) Python Dungeon Hunters Telegram -https://t.me/Coding_solution_0

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#include<bits/stdc++.h> using namespace std; #define ll long long #define MOD 998244353 #define add(x,y) ((x+y)%MOD) #define mul(x,y) ((x*y)%MOD) #define sub(x,y) ((MOD+x-y)%MOD) ll fastpow(ll x, ll y){ ll temp = 1; while(y){ if(y&1) temp = mul(temp, x); x = mul(x, x); y>>=1; } return temp; } ll inv(ll x){ return fastpow(x,MOD-2); } ll frac[200100]; ll C(ll n, ll i){ return mul(frac[n], inv(mul(frac[i], frac[n-i]))); } int main(){ ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); int n,k; cin>>n>>k; if(k>=n){ cout<<0<<endl; return 0; } ll ans = 0; frac[0] = 1; for(int i=1;i<=n;++i){ frac[i] = mul(i, frac[i-1]); } for(int i=0;i<=n-k;++i){ ll temp = mul(C(n-k,i), fastpow(n-k-i,n)); if(i&1) ans = sub(ans, temp); else ans = add(ans,temp); } ans = mul(ans, C(n, n-k)); if(k) ans = mul(ans, 2); cout<<ans<<endl; } CHESS GAME CODE Telegram - https://t.me/Coding_solution_0

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