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Athletes code Accenture Hackdiva

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Nodes code in python Accenture Hackdiva Code

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C

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LI.GCD code in another way of solving Infosys

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LI Gcd Code in python Infosys

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def f(S): n = len(S) lf, rf = {}, {} ls, rs = set(), set() # Initialize the rf map and rs with the entire string S for c in S: rf[c] = rf.get(c, 0) + 1 rs.add(c) mx = 0 # Traverse the string and adjust the ls and rs sets and maps for i in range(n - 1): c = S[i] lf[c] = lf.get(c, 0) + 1 rf[c] -= 1 if rf[c] == 0: rs.remove(c) ls.add(c) cs = len(ls) + len(rs) mx = max(mx, cs) return n - mx // spilit screen

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public static int solve(int N, int[] A) { int t = 0; for (int num : A) { t += num; } int x = 0; int y = 0; for (int i = 0; i < N; i++) { int res = t - x - A[i]; if (x == res) { y++; } x += A[i]; } return y; } //Equilibrium Point

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def GetAnswer(N, A, B, P): dp = [0] * (N + 1) max_d = 0 for i in range(N - 1, -1, -1): max_p = P[i] min_p = P[i] for j in range(1, B + 1): if i + j <= N: max_p = max(max_p, P[i + j - 1]) min_p = min(min_p, P[i + j - 1]) max_d = max(max_d, max_p - min_p) if i + 1 == N: dp[i] = max(dp[i], max_d) else: dp[i] = max(dp[i], max_d, dp[i + 1]) return dp[0] import sys input = sys.stdin.read data = input().split() N = int(data[0]) A = int(data[1]) B = int(data[2]) P = list(map(int, data[3:])) result = GetAnswer(N, A, B, P) print(result) // XOR formaating code

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def main(): import sys input = sys.stdin.read data = input().split('\n') S = data[0].strip() N = int(data[1].strip()) W = [] for i in range(N): W.append(data[2 + i].strip()) freqS = get_frequency(S) count = 0 for w in W: if is_valid_anagram_subsequence(freqS, w): count += 1 print(count) def get_frequency(S): freq = [0] * 26 for c in S: freq[ord(c) - ord('a')] += 1 return freq def is_valid_anagram_subsequence(freqS, w): freqW = [0] * 26 for c in w: freqW[ord(c) - ord('a')] += 1 for i in range(26): if freqW[i] > freqS[i]: return False return True if name == "main": main() // MInimal subarray length

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