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LI.GCD code in another way of solving

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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 @allcoding1_official

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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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