Next palindrome number || 17 aug potd gfg ||

[@LokeshSharma-hm5jz]

just watched the approach and code it myself... hats off, thanks for sharing the approach and for a very beautiful explanation, for each edge case

[@SCODESam]

Glad it was helpful!

[@kuldeep6055]

bhai everything is good but explanation thoda aur clear chaihiye tha , you should use hindi also to explain clearly , but thankyou so much

[@SCODESam]

solution ->
class Solution
{
public:
bool allnine(int num[], int n)
{
for (int i = 0; i < n; i++)
if (num[i] != 9)
return false;
return true;
}
vector generateNextPalindrome(int num[], int n)
{
if (allnine(num, n))
{
vector ans;
ans.push_back(1);
for (int i = 0; i < n - 1; i++)
{
ans.push_back(0);
}
ans.push_back(1);
return ans;
}
else if (n & 1)
{
bool flag = false;
for (int i = 0; i < n / 2; i++)
{
if (num[i] > num[n - i - 1])
{
flag = true;
}
else if ((num[i] == num[n - i - 1]))
{
continue;
}
else
{
flag = false;
}
num[n - i - 1] = num[i];
}
if (flag == false)
{
int j = 0;
for (; j < n / 2 - 1; j++)
{
if (num[n / 2 - j] == 9)
{
num[n / 2 - j] = 0;
num[n / 2 + j] = 0;
}
else
break;
}
if (j == 0)
{
num[n / 2 - j]++;
}
else
{
num[n / 2 - j]++;
num[n / 2 + j]++;
}
}
}
else
{
bool flag = false;
for (int i = 0; i < n / 2; i++)
{
if (num[i] > num[n - i - 1])
flag = true;
else if ((num[i] == num[n - i - 1]))
continue;
else
flag = false;
num[n - i - 1] = num[i];
}
if (flag == false)
{
int j = 0;
for (; j < n / 2 - 1; j++)
{
if (num[n / 2 - j - 1] == 9)
{
num[n / 2 - j - 1] = 0;
num[n / 2 + j] = 0;
}
else
break;
}
num[n / 2 - j - 1]++;
num[n / 2 + j]++;
}
}
vector ans(n);
for (int i = 0; i < n; i++)
{
ans[i] = num[i];
}
return ans;
}
};