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UkG Help available

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IBM all codes are done

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def getMinimumOperations(tree_nodes, tree_from, tree_to, initial, expected):
adj = [[] for _ in range(tree_nodes)]
for i in range(len(tree_from)):
adj[tree_from[i]].append(tree_to[i])
adj[tree_to[i]].append(tree_from[i])

parent = [-1] * tree_nodes
depth = [0] * tree_nodes
children = [[] for _ in range(tree_nodes)]

def dfs(node, par):
for next_node in adj[node]:
if next_node != par:
parent[next_node] = node
depth[next_node] = depth[node] + 1
children[node].append(next_node)
dfs(next_node, node)

dfs(0, -1)

def flip_subtree(curr_values, node):
new_values = curr_values.copy()
stack = [node]
while stack:
current = stack.pop()
if current % 2 == 0:
new_values[current] = 1 - new_values[current]
stack.extend(children[current])
return new_values

def get_path_to_root(node):
path = []
while node != -1:
path.append(node)
node = parent[node]
return path

operations = 0
current = initial.copy()

for node in range(tree_nodes):
if current[node] != expected[node]:
path = get_path_to_root(node)
found = False
for ancestor in path:
test_values = flip_subtree(current, ancestor)
if test_values[node] == expected[node]:
current = test_values
operations += 1
found = True
break
if not found:
return -1

return operations if current == expected else -1

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UkG || 2yrs EXPERIENCE Help done ✔️

Coding+ git repository
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#UkG #2yrs_Experience
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Wipro Elite Help available

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@Mrtrueliving_ix

@Mrtrueliving_ix

Previous Helping proofs

https://t.me/code_alphix/5153

https://t.me/code_alphix/5092?single


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Remember,Hard work in the wrong place is always  wasted 💯

                                   - @MRTRUELIVING_IX
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Join the Wipro discussion group

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💯 Free coding Help
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def solution(S, K):
n = len(S)
res = float('inf')
for i in range(n - K + 1):
t = S[:i] + S[i + K:]
compressed = ''
count = 1
for j in range(1, len(t)):
if t[j] == t[j - 1]:
count += 1
else:
compressed += t[j - 1] + (str(count) if count > 1 else '')
count = 1
if t:
compressed += t[-1] + (str(count) if count > 1 else '')
res = min(res, len(compressed))
return res

Microsoft ✔️
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Proof ✔️
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def solution(s, t):
n = len(s)
sw = s[0] > t[0]
c = 0
for k in range(n):
if s[k] != t[k]:
if (s[k] > t[k]) == sw:
c += 1
return c


Microsoft ✔️
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Join the Wipro discussion group

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💯 Free coding Help.....✔️
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All set for Wipro 12 pm slot ✔️...

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@Mrtrueliving_ix ✔️
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Wipro 12 pm || Offcampus Help done

Test accomplished


Slot - 1


Those who need Test Clearance 💯

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@Mrtrueliving_ix @Mrtrueliving_ix

#WiproElite #offCAMPUS #Round1
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