Sry for the above typo. Correct algo
A more efficient approach :
Suppose the array is M*N
fun(int i, int j)
if(a[i][j] == x)
return;
mid1=(i+M-1)/2;
mid2=(j+N-1)/2;
if(abs(a[i][mid1] - x) < abs(a[mid2][j]) - x)
return fun(i,mid1)
else
return fun(mid2,j)
This algo can be easily modified to account for the case in which the
element does not exist
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