Minimum length Unsorted Subarray, sorting which makes the array sorted | GeeksforGeeks

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[snipegams]

He's so invested in reading it out that he even pronounced changed to chang-ed. Wow whatever the reason seems to be feel like we should empower more videos like this

[SachinSingh-lbqz]

with space complexity can have better solution it keep two array of min and max element till any index from l to r and r to L?

[wecan]

class Solution{
public:
vector<int> printUnsorted(int arr[], int n)
{
int s, e, max, min;
vector<int>v;
// step 1(a) of above algo
for (s = 0; s < n-1; s++)
{
if (arr[s] > arr[s+1])
break;
}
if (s == n-1)
{
v.push_back(0);
v.push_back(0);
return v;
}

// step 1(b) of above algo
for(e = n - 1; e > 0; e--)
{
if(arr[e] < arr[e-1])
break;
}

// step 2(a) of above algo
max = arr[s]; min = arr[s];
for(int i = s + 1; i <= e; i++)
{
if(arr[i] > max)
max = arr[i];
if(arr[i] < min)
min = arr[i];
}

// step 2(b) of above algo
for( int i = 0; i < s; i++)
{
if(arr[i] > min)
{
s = i;
break;
}
}

// step 2(c) of above algo
for( int i = n -1; i >= e+1; i--)
{
if(arr[i] < max)
{
e = i;
break;
}
}

v.push_back(s);
v.push_back(e);
return v;
}
};

[guruprakash]

Thanks for reading the solution, I cannot read texts.

[abijeetshyam]

Is this called expaination or reading ppts? I wonder!

[siddharthachatterjee]

Fails for arrays with repeated element.
A : [ 1, 1, 10, 10, 15, 10, 15, 10, 10, 15, 10, 15 ]