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Netcore Help βœ…


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πŸ“’ Accenture ONCAMPUS 2026 Test Pattern


Round 1: Aptitude / Personal / Logical


1️⃣ Personal Feedback – 54 questions (psychometric / behavioral)

2️⃣ Route the Plan – 3 questions (logic / flow-based reasoning)

3️⃣ Select Low to High Value – 25 questions
4️⃣ Escape Room – 3 questions (advanced logic / puzzle)

Round 2: Technical MCQs


πŸ“ 45 questions | 4-option MCQs
Topics: Programming, DBMS, OOP, Networking, Data Structures, Algorithms

Round 3: Coding & Simulation


πŸ’» Coding Question – Python / Java / C++

πŸ—„οΈ SQL Question – Queries on a given DB schema

🌐 Simulation – HTML, CSS, JS (UI / DOM logic)

πŸ”₯ Preparation Tip:

Focus on logic puzzles and coding practice

Revise database queries and front-end fundamentals

Be ready for behavioral/psychometric assessment


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πŸ’» Placement Practice: Netcore

def solve(N, K, L, R, Sn, Sk):
s, pre, res = set(Sk), [0]*(N+1), 0
for i in range(N): pre[i+1] = pre[i] + (Sn[i] in s)
j1 = j2 = 0
for i in range(N):
while j1 < N and pre[j1+1]-pre[i] < L: j1 += 1
while j2 < N and pre[j2+1]-pre[i] <= R: j2 += 1
res += max(0, j2 - j1)
return res

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Netcore Help βœ…


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#Netcore #Offcampus
def solve(n, m, s, r, q, x, y):
M = 998244353
def pw(a, b):
r1 = 1
while b:
if b & 1: r1 = r1 * a % M
a = a * a % M
b >>= 1
return r1
f = [1] * (n + 1)
for i in range(1, n + 1):
f[i] = f[i - 1] * i % M
iv = [1] * (n + 1)
iv[n] = pw(f[n], M - 2)
for i in range(n, 0, -1):
iv[i - 1] = iv[i] * i % M
def C(a, b):
return 0 if b < 0 or b > a else f[a] * iv[b] % M * iv[a - b] % M
rc = [0] * 26
for c in r: rc[ord(c) - 97] += 1
id = [i for i in range(26) if rc[i]]
p = [[0] * (n + 1) for _ in range(26)]
for i in range(1, n + 1):
ch = ord(s[i - 1]) - 97
for c in range(26): p[c][i] = p[c][i - 1]
p[ch][i] += 1
ans = []
for i in range(q):
l, rr = x[i], y[i]
if rr - l + 1 < m: ans.append(0); continue
ok, v = 1, 1
for c in id:
hv = p[c][rr] - p[c][l - 1]
nd = rc[c]
if hv < nd: ok = 0; break
v = v * C(hv, nd) % M
ans.append(v if ok else 0)
return ans

Netcore βœ“
def find_maximum_remainder(N, A, k):
    from bisect import bisect_left
    e = sorted(list(set(x % k for x in A if x % 2 == 0)))
    o = sorted(list(set(x % k for x in A if x % 2 != 0)))
    ans = 0
    for i in e:
        q = bisect_left(o, (k-i) % k)
        if q < len(o) and (i + o[q]) % k > ans:
            ans = max(ans, (i + o[q]) % k, (i + o[-1]) % k)
        if q > 0: ans = max(ans, (i + o[q-1]) % k, (i + o[-1]) % k)
    return ans

Netcore βœ“
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