The approach is to find the right smaller and left smaller element and find the largest Rectangle area in Histogram.

This is done by taking different bars and finding the maximum width possible using the bar.

Dry Run

Considering the width of each bar as 1 unit.

For first bar, area possible = 2* 1 =2 sq.units

For second bar, area possible = 1 * 6 =6 sq.units

For third bar , area possible = 5 *2 = 10 sq.units

For fourth bar , area possible = 6 * 1 = 6 sq.units

For Fifth bar , area possible = 2 * 4 = 8 sq.units

For Sixth bar , area possible = 3 * 1 =3 sq.units

So, the maximum area possible = 10 sq units.