A more efficient approach :
Suppose the array is M*N
i=j=0;
if(a[i][j] == x)
return;
mid1=(i+M-1)/2;
mid2=(j+N-1)/2;
if(a[i][mid1]==x || a[mid2][j])==x
if(abs(a[i][mid1] - x) < abs(a[mid2][j]) - x)
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